Solving Using the Quadratic Formula

What Is the Quadratic Formula?

The quadratic formula is a method for solving any quadratic equation written in standard form:

ax² + bx + c = 0

where:

a ≠ 0

The formula gives the possible values of x:

x = [−b ± √(b² − 4ac)] / (2a)

Unlike factoring, which is easiest when a quadratic has simple factors, the quadratic formula can be used for any quadratic equation.

It can produce:

• two different real solutions
• one repeated real solution
• no real solutions
• irrational solutions
• fractional solutions

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Identifying a, b and c

Before using the quadratic formula, the equation must be written in standard form:

ax² + bx + c = 0

Then identify:

• a = coefficient of x²
• b = coefficient of x
• c = constant term

For example:

2x² − 7x + 3 = 0

gives:

a = 2

b = −7

c = 3

The signs are part of the coefficients.

This is especially important when b or c is negative.

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More Examples of Identifying Coefficients

If a term is missing, its coefficient is zero.

For:

x² − 16 = 0

there is no x-term, so:

b = 0

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Put the Equation Equal to Zero First

Before identifying a, b and c, rearrange the equation into:

ax² + bx + c = 0

For example:

x² + 3x = 10

should first become:

x² + 3x − 10 = 0

Now:

a = 1

b = 3

c = −10

Identifying the coefficients before rearranging can lead to incorrect substitution.

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Understanding the Formula

The quadratic formula contains several important parts:

−b

means the opposite of b.

±

means there are potentially two solutions:

• one using +
• one using −

b² − 4ac

is called the discriminant.

The denominator:

2a

applies to the entire numerator.

A useful way to think about the formula is:

x = (entire numerator) / (2a)

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A Reliable Method

When using the quadratic formula:

Write the equation in standard form

ax² + bx + c = 0

Identify the coefficients

Write:

a = ...

b = ...

c = ...

Substitute into the formula

Use brackets around negative numbers.

Calculate the discriminant

Find:

b² − 4ac

Calculate the square root

Simplify it exactly if possible.

Calculate both solutions

Use both + and −.

Simplify

Reduce fractions and radicals where possible.

Verify

Substitute the solutions into the original equation.

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Basic Example

Solve:

x² − 5x + 6 = 0

Identify:

a = 1

b = −5

c = 6

Substitute:

x = [−(−5) ± √((−5)² − 4(1)(6))] / [2(1)]

Simplify:

x = [5 ± √(25 − 24)] / 2

x = [5 ± √1] / 2

x = (5 ± 1)/2

Now calculate both possibilities.

Using +:

x = (5 + 1)/2

x = 6/2

x = 3

Using −:

x = (5 − 1)/2

x = 4/2

x = 2

Therefore:

x = 2 or x = 3

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Why the ± Symbol Matters

The formula:

x = [−b ± √(b² − 4ac)]/(2a)

actually represents two calculations.

One is:

x = [−b + √(b² − 4ac)]/(2a)

The other is:

x = [−b − √(b² − 4ac)]/(2a)

A quadratic can have two different solutions because a parabola can cross the x-axis at two different points.

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Example with a Leading Coefficient

Solve:

2x² − 7x + 3 = 0

Identify:

a = 2

b = −7

c = 3

Substitute:

x = [−(−7) ± √((−7)² − 4(2)(3))] / [2(2)]

Simplify:

x = [7 ± √(49 − 24)] / 4

x = [7 ± √25] / 4

x = (7 ± 5)/4

First solution:

x = (7 + 5)/4

x = 12/4

x = 3

Second solution:

x = (7 − 5)/4

x = 2/4

x = 1/2

Therefore:

x = 3 or x = 1/2

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Be Careful with Negative b

Consider:

3x² − 4x − 2 = 0

Here:

a = 3

b = −4

c = −2

Notice:

−b = −(−4) = 4

Also:

b² = (−4)² = 16

Using brackets around negative values helps prevent sign errors.

Substitute:

x = [4 ± √(16 − 4(3)(−2))] / 6

x = [4 ± √(16 + 24)] / 6

x = [4 ± √40] / 6

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Simplifying Radicals

The answer:

x = [4 ± √40]/6

can be simplified.

Since:

40 = 4 × 10

we have:

√40 = √4 × √10

√40 = 2√10

Therefore:

x = [4 ± 2√10]/6

Every term can be divided by 2:

x = [2 ± √10]/3

This is the simplified exact solution.

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Exact Solutions

An exact solution keeps square roots rather than replacing them with rounded decimals.

For example:

x = 3 ± √5

is exact.

Using:

√5 ≈ 2.236

gives approximate solutions:

x ≈ 5.236

and:

x ≈ 0.764

Unless a decimal answer is specifically needed, exact solutions are usually preferred.

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Simplifying Square Roots

Look for a perfect-square factor.

For example:

√12

12 = 4 × 3

so:

√12 = 2√3

Similarly:

√18 = 3√2

√20 = 2√5

√27 = 3√3

√32 = 4√2

√45 = 3√5

√48 = 4√3

√50 = 5√2

√72 = 6√2

Simplifying radicals is an important part of giving exact quadratic-formula solutions.

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Example with Irrational Solutions

Solve:

x² − 4x − 1 = 0

Identify:

a = 1

b = −4

c = −1

Substitute:

x = [4 ± √(16 − 4(1)(−1))]/2

x = [4 ± √20]/2

Simplify:

√20 = 2√5

So:

x = [4 ± 2√5]/2

Therefore:

x = 2 ± √5

These are the exact solutions.

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Approximate Solutions

From:

x = 2 ± √5

and:

√5 ≈ 2.236

we get:

x ≈ 4.236

and:

x ≈ −0.236

Exact form:

2 ± √5

Approximate form:

4.236 and −0.236

Both can be useful depending on the problem.

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The Discriminant

The expression under the square root is:

b² − 4ac

This is called the discriminant.

It is often represented by:

D = b² − 4ac

The discriminant tells us how many real solutions a quadratic has.

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When the Discriminant Is Positive

If:

D > 0

then √D is positive and the ± produces two different values.

Therefore, the equation has:

two distinct real solutions

For example:

x² − 5x + 6 = 0

D = (−5)² − 4(1)(6)

D = 25 − 24

D = 1

Since D > 0, there are two real solutions.

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When the Discriminant Is Zero

If:

D = 0

then:

√D = 0

The + and − calculations become identical.

Therefore, there is:

one repeated real solution

Consider:

x² − 6x + 9 = 0

D = (−6)² − 4(1)(9)

D = 36 − 36

D = 0

The formula gives:

x = [6 ± 0]/2

x = 3

This is a repeated root.

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When the Discriminant Is Negative

If:

D < 0

then the formula requires the square root of a negative number.

There are therefore:

no real solutions

For example:

x² + 2x + 5 = 0

D = 2² − 4(1)(5)

D = 4 − 20

D = −16

Since D < 0, there are no real roots.

The corresponding parabola does not cross the x-axis.

If complex numbers are being used, the equation can be solved further.

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Discriminant Summary

This allows us to predict the number of real roots without fully solving the equation.

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Example: Predict Before Solving

Consider:

2x² + 3x + 7 = 0

Calculate:

D = 3² − 4(2)(7)

D = 9 − 56

D = −47

We can immediately conclude:

There are no real solutions.

There is no need to continue if the question only asks for the number of real solutions.

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Example with a Missing x-Term

Solve:

2x² − 18 = 0

Here:

a = 2

b = 0

c = −18

Substitute:

x = [−0 ± √(0² − 4(2)(−18))]/[2(2)]

x = [±√144]/4

x = ±12/4

Therefore:

x = ±3

This equation could be solved more quickly using square roots, but the quadratic formula still works.

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Example with a Missing Constant

Solve:

3x² − 12x = 0

Here:

a = 3

b = −12

c = 0

Substitute:

x = [12 ± √144]/6

x = [12 ± 12]/6

Therefore:

x = 24/6 = 4

or:

x = 0/6 = 0

So:

x = 0 or x = 4

Again, factoring would be faster:

3x(x − 4) = 0

This shows that the quadratic formula works even when another method may be more efficient.

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Rearranging Before Using the Formula

Solve:

x² + 8 = 6x

First rearrange:

x² − 6x + 8 = 0

Now:

a = 1

b = −6

c = 8

Substitute:

x = [6 ± √(36 − 32)]/2

x = [6 ± 2]/2

Therefore:

x = 4

or:

x = 2

Always identify a, b and c after rearranging into standard form.

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Verifying Solutions

Suppose we solve:

2x² − 7x + 3 = 0

and obtain:

x = 3

and:

x = 1/2

Check x = 3:

2(3²) − 7(3) + 3

= 18 − 21 + 3

= 0

So x = 3 works.

Now check x = 1/2:

2(1/2)² − 7(1/2) + 3

= 2(1/4) − 7/2 + 3

= 1/2 − 7/2 + 3

= −3 + 3

= 0

Both solutions are verified.

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Verifying Irrational Solutions

Consider:

x² − 4x − 1 = 0

We found:

x = 2 ± √5

Check:

x = 2 + √5

Substitute:

(2 + √5)² − 4(2 + √5) − 1

Expand:

4 + 4√5 + 5 − 8 − 4√5 − 1

= 9 − 9

= 0

Therefore:

x = 2 + √5

is correct.

The same process verifies:

x = 2 − √5

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Using Decimal Verification

If the exact solution is:

x = 2 + √5

we can approximate:

x ≈ 4.236

Substituting 4.236 into:

x² − 4x − 1

gives a value very close to zero.

It may not equal exactly zero because the value of x has been rounded.

For exact verification, use the exact radical form.

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Quadratic Formula vs Factoring

Consider:

x² − 5x + 6 = 0

This factors immediately:

(x − 2)(x − 3) = 0

Factoring is probably the faster method.

But consider:

x² − 4x − 1 = 0

This does not factor neatly using integers.

The quadratic formula gives:

x = 2 ± √5

In this case, the quadratic formula is more convenient.

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Quadratic Formula vs Completing the Square

Both methods can solve quadratics that do not factor easily.

For example:

x² − 6x + 2 = 0

Completing the square gives:

x = 3 ± √7

The quadratic formula gives the same answer.

The quadratic formula is often faster for direct solving, while completing the square is especially useful when you also want to understand the vertex form of the quadratic.

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Choosing a Solving Method

There may be several correct methods for the same equation.

The goal is to choose one that is accurate and efficient.

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Why the Quadratic Formula Works

The quadratic formula is not an arbitrary rule.

It can be derived from:

ax² + bx + c = 0

using completing the square.

This is why the formula works for all quadratic equations.

The quantities a, b and c contain all the information needed to determine the roots.

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Connection to the Graph

The solutions of:

ax² + bx + c = 0

are the x-values where:

y = ax² + bx + c

has:

y = 0

Therefore, the solutions found using the quadratic formula correspond to the x-intercepts of the parabola.

If there are two real solutions, the graph crosses the x-axis twice.

If there is one repeated solution, the graph touches the x-axis once.

If there are no real solutions, the graph does not meet the x-axis.

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Worked Example: Complete Process

Solve:

3x² + 5x − 2 = 0

Identify:

a = 3

b = 5

c = −2

Substitute:

x = [−5 ± √(5² − 4(3)(−2))]/[2(3)]

Simplify:

x = [−5 ± √(25 + 24)]/6

x = [−5 ± √49]/6

x = (−5 ± 7)/6

First solution:

x = (−5 + 7)/6

x = 2/6

x = 1/3

Second solution:

x = (−5 − 7)/6

x = −12/6

x = −2

Therefore:

x = 1/3 or x = −2

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Worked Example: Exact Radical Answer

Solve:

2x² + 4x − 3 = 0

Identify:

a = 2

b = 4

c = −3

Substitute:

x = [−4 ± √(4² − 4(2)(−3))]/4

x = [−4 ± √(16 + 24)]/4

x = [−4 ± √40]/4

Simplify:

√40 = 2√10

So:

x = [−4 ± 2√10]/4

Divide numerator and denominator by 2:

x = [−2 ± √10]/2

These are the exact solutions.

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Worked Example: No Real Solutions

Solve over the real numbers:

x² + 4x + 8 = 0

a = 1

b = 4

c = 8

Discriminant:

D = 4² − 4(1)(8)

D = 16 − 32

D = −16

Because:

D < 0

the equation has:

no real solutions

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Worked Example: Repeated Root

Solve:

4x² − 12x + 9 = 0

a = 4

b = −12

c = 9

Discriminant:

D = (−12)² − 4(4)(9)

D = 144 − 144

D = 0

Substitute:

x = [12 ± 0]/8

Therefore:

x = 3/2

This is a repeated root.

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Common Mistakes

Mistake: Identifying coefficients before rearranging the equation.
First write the equation as ax² + bx + c = 0.

Mistake: Forgetting that the sign belongs to the coefficient.
For 2x² − 7x + 3 = 0, b = −7, not 7.

Mistake: Calculating −b incorrectly.
If b = −7, then −b = 7.

Mistake: Squaring a negative number without brackets.
If b = −5, use (−5)² = 25.

Mistake: Forgetting ±.
Most quadratics with a positive discriminant have two solutions.

Mistake: Dividing only the square-root term by 2a.
The entire numerator is divided by 2a.

Mistake: Entering the formula incorrectly into a calculator.
Use brackets around the complete numerator and denominator.

Mistake: Leaving a radical unsimplified.
For example, √20 should normally be simplified to 2√5.

Mistake: Rounding too early.
Keep exact values until the final step whenever possible.

Mistake: Not checking the solutions.
Substitution into the original equation can reveal sign or arithmetic errors.

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Key Terms

Quadratic equation — equation that can be written as ax² + bx + c = 0, where a ≠ 0.

Coefficient — numerical value multiplying a variable or power of a variable.

Standard form — ax² + bx + c = 0.

Quadratic formula — general formula used to calculate the roots of a quadratic equation.

Root — value of x that makes the quadratic equal to zero.

Solution — value that makes an equation true.

Discriminant — expression b² − 4ac that determines the nature and number of real roots.

Radical — expression containing a root symbol, such as √10.

Exact solution — solution left in an exact form such as 2 + √5.

Approximate solution — decimal approximation of an exact value.

Repeated root — a root that occurs twice.

Real solution — solution belonging to the real number system.

Verification — checking a solution by substituting it into the original equation.

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Key Takeaways

• The quadratic formula solves equations written as ax² + bx + c = 0.
• Always rearrange the equation into standard form before identifying the coefficients.
• The coefficient a belongs to x², b belongs to x, and c is the constant.
• Include negative signs when identifying coefficients.
• The quadratic formula is x = [−b ± √(b² − 4ac)]/(2a).
• Use brackets when substituting negative coefficients.
• The ± symbol usually produces two possible solutions.
• The expression b² − 4ac is the discriminant.
• A positive discriminant gives two distinct real roots.
• A zero discriminant gives one repeated real root.
• A negative discriminant gives no real roots.
• Exact solutions should usually be simplified before converting them to decimals.
• Radicals such as √40 can be simplified by finding perfect-square factors.
• The quadratic formula works even when a quadratic cannot be factored easily.
• Factoring may still be faster when simple factors are obvious.
• Completing the square and the quadratic formula can produce the same exact solutions.
• The roots found by the quadratic formula correspond to the x-intercepts of the parabola.
• Substitution into the original equation can be used to verify solutions.

Check Your Understanding

1. Identify a, b and c in 3x² − 7x + 2 = 0.
2. Identify a, b and c in −2x² + 5x − 8 = 0.
3. Identify a, b and c in 4x² − 12 = 0.
4. Rearrange x² + 8 = 5x into standard form and identify a, b and c.
5. Solve x² − 7x + 10 = 0 using the quadratic formula.
6. Solve 2x² − 5x − 3 = 0 using the quadratic formula.
7. Solve 3x² + 7x + 2 = 0 using the quadratic formula.
8. Solve x² − 6x + 1 = 0 and give exact solutions.
9. Solve 2x² + 4x − 5 = 0 and simplify the exact solutions.
10. Solve x² + 4x + 8 = 0 over the real numbers.
11. Determine the discriminant of x² − 10x + 25 = 0. What does it tell you?
12. Determine the number of real roots of 3x² + 2x + 5 = 0 without fully solving.
13. Simplify √28.
14. Simplify √45.
15. Explain why brackets should be used when substituting a negative value for b.
16. Explain why the ± symbol is necessary in the quadratic formula.
17. Explain what the discriminant tells us about the graph of a quadratic.
18. Solve 4x² − 4x − 3 = 0 and verify both solutions by substitution.
19. Give an example of a quadratic for which factoring would probably be faster than using the quadratic formula.
20. Explain why the quadratic formula is useful even when other quadratic-solving methods are available.