The Quadratic Formula

What the quadratic formula does

A quadratic equation is any equation that can be written as:

ax² + bx + c = 0

where a, b and c are numbers and a ≠ 0. The quadratic formula is a single tool that solves every such equation:

x = (−b ± √(b² − 4ac)) / (2a)

Unlike factoring, which only works when the numbers cooperate, the formula always works. If you want the bigger picture of all solving methods, see the lesson on quadratic equations; this lesson focuses on using the formula confidently and correctly.

Read it aloud as: "x equals negative b, plus or minus the square root of b squared minus four a c, all over two a." Saying it this way helps you remember that the whole top is divided by 2a, not just part of it.

Step 1: Get the equation into standard form

Before reading off a, b and c, the equation must equal zero on the right and be in the order ax² + bx + c = 0.

Worked example 1: Rearrange x² + 5 = 6x.

1. Subtract 6x from both sides: x² − 6x + 5 = 0.
2. Now it is in standard form, so a = 1, b = −6, c = 5.

The signs matter. Here b = −6, not 6. Getting the signs of a, b, c right is where most mistakes happen, so always include the sign that sits in front of each term.

Step 2: Substitute carefully

Once you have a, b and c, put them into the formula. Use brackets around every value to protect the signs.

Worked example 2: Solve 2x² + 3x − 2 = 0.

1. Identify: a = 2, b = 3, c = −2.
2. Discriminant: b² − 4ac = (3)² − 4(2)(−2) = 9 + 16 = 25.
3. Square root: √25 = 5.
4. Substitute: x = (−3 ± 5) / (2 × 2) = (−3 ± 5) / 4.
5. Plus case: x = (−3 + 5)/4 = 2/4 = 0.5.
6. Minus case: x = (−3 − 5)/4 = −8/4 = −2.
7. Solutions: x = 0.5 or x = −2.

Check: Substitute x = −2 into the original: 2(−2)² + 3(−2) − 2 = 8 − 6 − 2 = 0. Correct.

The discriminant: a preview of the answer

The expression b² − 4ac under the root is called the discriminant, often written as Δ (delta). Calculating it first tells you what kind of answer to expect.

Discriminant b² − 4ac	Real solutions	What the parabola does
Positive (> 0)	Two different roots	Crosses the x-axis twice
Zero (= 0)	One repeated root	Just touches the x-axis
Negative (< 0)	No real roots	Floats above (or below) the axis

Worked example 3: How many real solutions does x² + 2x + 5 = 0 have?

1. a = 1, b = 2, c = 5.
2. Discriminant: 2² − 4(1)(5) = 4 − 20 = −16.
3. Negative, so there are no real solutions. There is no point continuing with the formula.

When the answer is a surd

Often the discriminant is not a perfect square. Then leave the answer in exact surd form (review surds if needed), or round to a sensible decimal.

Worked example 4: Solve x² − 4x + 1 = 0.

1. a = 1, b = −4, c = 1.
2. Discriminant: (−4)² − 4(1)(1) = 16 − 4 = 12.
3. √12 = 2√3 (since 12 = 4 × 3).
4. Substitute: x = (4 ± 2√3) / 2.
5. Divide every term on top by 2: x = 2 ± √3.
6. Exact solutions: x = 2 + √3 or x = 2 − √3 (about 3.73 and 0.27).

Worked example 5: Solve 3x² − 5x − 1 = 0, to two decimal places.

1. a = 3, b = −5, c = −1.
2. Discriminant: (−5)² − 4(3)(−1) = 25 + 12 = 37.
3. √37 ≈ 6.08.
4. x = (5 ± 6.08) / 6.
5. Plus: x ≈ 11.08/6 ≈ 1.85. Minus: x ≈ −1.08/6 ≈ −0.18.
6. Solutions: x ≈ 1.85 or x ≈ −0.18.

A worked word problem

The formula really earns its keep on real questions where factoring is hopeless.

Worked example 6: A ball is thrown upward. Its height in metres after t seconds is h = 20t − 5t². When does it hit the ground (height 0)?

1. Set h = 0: 20t − 5t² = 0, i.e. −5t² + 20t = 0. Multiply through by −1: 5t² − 20t = 0.
2. Here a = 5, b = −20, c = 0.
3. Discriminant: (−20)² − 4(5)(0) = 400 − 0 = 400, and √400 = 20.
4. t = (20 ± 20) / (2 × 5) = (20 ± 20)/10.
5. Plus case: t = 40/10 = 4. Minus case: t = 0/10 = 0.
6. t = 0 is the launch moment; t = 4 seconds is when it lands.

This shows a useful habit: even when c = 0, the formula still works, and the two roots both carry meaning.

A common mistake to avoid

A frequent slip is dividing only part of the numerator by 2a. The formula divides the entire top — both −b and the root — by 2a. Another is mishandling −4ac when c is negative: −4 × a × (−c) becomes positive. In Worked example 2, −4(2)(−2) = +16, which is why the discriminant grew rather than shrank. Writing every value in brackets prevents both errors.

A repeated root

Worked example 7: Solve x² + 6x + 9 = 0.

1. a = 1, b = 6, c = 9.
2. Discriminant: 6² − 4(1)(9) = 36 − 36 = 0.
3. √0 = 0, so the ± adds nothing.
4. x = (−6 ± 0)/2 = −3.
5. One repeated solution: x = −3. The parabola just touches the x-axis here.

Where the quadratic formula is used

Any situation that curves or accelerates leads to a quadratic. Physicists use it to find when a thrown ball lands (height = 0). Engineers solve quadratics to size beams and arches. In business, the price that maximises revenue is found from a quadratic. Because real-world numbers are rarely tidy, the quadratic formula — not factoring — is the everyday workhorse. To see how these curves look when plotted, visit functions and graphs.

Why it works: The formula is not magic — it is the result of completing the square on the general equation ax² + bx + c = 0. Doing that algebra once, with letters instead of numbers, produces a formula that then works for all numbers at once.

Practice activity

Solve each using the quadratic formula. Give exact answers, or round to 2 d.p. where needed.

1. x² − 7x + 12 = 0
2. x² + 2x − 1 = 0
3. 2x² + 5x − 3 = 0
4. x² − 6x + 9 = 0
5. x² + x + 1 = 0 (look at the discriminant first)

Answers:

1. a=1, b=−7, c=12. Disc = 49 − 48 = 1, √1 = 1. x = (7 ± 1)/2 → x = 4 or x = 3.
2. a=1, b=2, c=−1. Disc = 4 + 4 = 8, √8 = 2√2. x = (−2 ± 2√2)/2 → x = −1 ± √2.
3. a=2, b=5, c=−3. Disc = 25 + 24 = 49, √49 = 7. x = (−5 ± 7)/4 → x = 0.5 or x = −3.
4. a=1, b=−6, c=9. Disc = 36 − 36 = 0. x = 3 (repeated root).
5. a=1, b=1, c=1. Disc = 1 − 4 = −3, which is negative → no real solutions.

Summary

The quadratic formula x = (−b ± √(b² − 4ac)) / (2a) solves any quadratic in standard form. The reliable routine is: rewrite as ax² + bx + c = 0, read off a, b, c with their signs, compute the discriminant b² − 4ac to predict the number of roots, then substitute and use the ± to get both answers. Keep solutions exact in surd form unless a decimal is requested, and always check by substituting back.