The Quadratic Formula

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Skill: quadratic-formula

The Quadratic Formula

The quadratic formula solves every quadratic equation ax² + bx + c = 0 — even the ones that refuse to factor. Memorize it once, use it forever: it is the safety net underneath every other solving method.

1 The Formula

For any quadratic equation in standard form ax² + bx + c = 0 (with a ≠ 0), the solutions are given by one formula:

x = −b ± √b² − 4ac2a

The part under the radical, b² − 4ac, is called the discriminant — it is a fortune-teller. Before you finish the arithmetic, it already tells you what kind of answers to expect:

  • Discriminant positive → two real solutions (the parabola crosses the x-axis twice).
  • Discriminant zero → exactly one solution, a “double root” (the parabola just touches the x-axis).
  • Discriminant negative → no real solutions (the parabola never touches the x-axis — stop here, no need to force a square root of a negative).
x = -1x = 4two real solutionsD = 25x = 3one solution (double root)D = 0no real solutionsD = −4
The discriminant predicts the picture before you compute: positive → two crossings, zero → one touch, negative → no touch. Formula and graph always agree.

2 Step Zero: Label a, b, c

Before touching the formula, label the coefficients — signs included. Most formula errors are labeling errors, not arithmetic errors.

In x² + 5x + 6 = 0, the number attached to x² is a = 1, the number attached to x is b = 5, and the constant term is c = 6.

Then substitute with the colors matching:

x = −(5) ± √(5)² − 4(1)(6)2(1)

Watch the minus on b. If b = −7, then −b = −(−7) = +7. Write b with its sign first, then compute −b as a separate step.

x² + 5x + 6 = 0 a = 1 b = 5 c = 6 coefficient of x² · coefficient of x · constant
Label first, substitute second. The colors follow the numbers into the formula.

3 Worked Examples

The pattern never changes: label a, b, c → compute the discriminant → substitute → simplify → check by substitution.

Example 1 Basic — solve x² + 5x + 6 = 0
  1. Label: a = 1, b = 5, c = 6.
  2. Discriminant: D = b² − 4ac = 25 − 4(1)(6) = 25 − 24 = 1 (positive → two solutions).
  3. Substitute: x = −5 ± √12 = −5 ± 12.
  4. Two values: x = −5 + 12 = −2 or x = −5 − 12 = −3.
x = −2 or x = −3. Check: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 ✓; (−3)² + 5(−3) + 6 = 9 − 15 + 6 = 0 ✓.
Example 2 With a leading coefficient — solve 2x² − 7x + 3 = 0
  1. Label carefully: a = 2, b = −7, c = 3. So −b = −(−7) = 7.
  2. Discriminant: D = (−7)² − 4(2)(3) = 49 − 24 = 25 (positive → two solutions).
  3. Substitute: x = 7 ± √254 = 7 ± 54.
  4. Two values: x = 124 = 3 or x = 24 = 12.
x = 3 or x = 1/2. Check: 2(9) − 7(3) + 3 = 18 − 21 + 3 = 0 ✓; 2(1/4) − 7(1/2) + 3 = 0.5 − 3.5 + 3 = 0 ✓.
Example 3 Discriminant zero — solve x² − 6x + 9 = 0
  1. Label: a = 1, b = −6, c = 9. So −b = 6.
  2. Discriminant: D = (−6)² − 4(1)(9) = 36 − 36 = 0 → exactly one solution (the ± collapses).
  3. Substitute: x = 6 ± √02 = 62 = 3.
x = 3 (double root). Check: 9 − 18 + 9 = 0 ✓. The graph touches the x-axis once at x = 3.
Example 4 Negative discriminant — solve x² + 4x + 5 = 0
  1. Label: a = 1, b = 4, c = 5.
  2. Discriminant: D = 16 − 4(1)(5) = 16 − 20 = −4.
  3. Negative → no real solutions. Stop here — there is no need to force a square root of a negative number.
No real solutions. The parabola y = x² + 4x + 5 has its vertex at (−2, 1) — it floats above the x-axis and never crosses.

4 Common Mistakes

Three errors that show up on nearly every quadratic-formula quiz. Spot them now and they will never cost you points.

Mistake 1: Sign errors on −b
Wrong
For 2x² − 7x + 3 = 0, writing x = −7 ± …4 — forgetting that −b with b = −7 gives +7.
Right
x = −(−7) ± √254 = 7 ± 54.
Fix: write b with its sign FIRST (b = −7), then compute −b as a separate step. Two negatives make a positive — every time.
Mistake 2: Dividing only part of the numerator by 2a
Wrong
x = −b ± √252a — the ± term escapes the division.
Right
x = −b ± √b² − 4ac2a — the ENTIRE top is divided by 2a.
Memory hook: draw the fraction bar long and put parentheses around 2a. The bar is a roof — everything under it gets divided.
Mistake 3: Arithmetic slips in the discriminant
Wrong
For 3x² + 5x + 2 = 0: computing 4ac as 4·3 = 12, so D = 25 − 12 = 13 — forgetting to multiply by c too.
Right
4ac = 4 · 3 · 2 = 24, so D = 25 − 24 = 1.
Fix: compute 4ac as its own mini-step: 4 × a × c, then subtract. And always ask: does the discriminant’s sign match the graph I expect?

5 Quick Check

Try each one on paper first, then reveal the answer. Always check by substituting back.

1. Solve x² − 3x − 4 = 0 using the quadratic formula.
Answer
Label: a = 1, b = −3, c = −4. Discriminant: (−3)² − 4(1)(−4) = 9 + 16 = 25.
x = 3 ± √252 = 3 ± 52
x = 4 or x = −1. Check: 16 − 12 − 4 = 0 ✓; 1 + 3 − 4 = 0 ✓.
2. Solve x² + 7x + 10 = 0 using the quadratic formula.
Answer
Discriminant: 49 − 40 = 9.
x = −7 ± √92 = −7 ± 32
x = −2 or x = −5. Check: 4 − 14 + 10 = 0 ✓; 25 − 35 + 10 = 0 ✓.
3. Solve 2x² − 12x + 18 = 0. (Hint: check the discriminant first.)
Answer
Discriminant: (−12)² − 4(2)(18) = 144 − 144 = 0 → exactly one solution.
x = 12 ± √04 = 124 = 3
x = 3 (double root). Check: 2(9) − 12(3) + 18 = 18 − 36 + 18 = 0 ✓.

Key Points to Remember

  • x = −b ± √b² − 4ac2a solves every ax² + bx + c = 0.
  • The discriminant D = b² − 4ac: positive → two real solutions; zero → one double root; negative → no real solutions.
  • Label a, b, c with their signs FIRST — most errors are labeling errors.
  • The whole numerator is divided by 2a — draw the fraction bar long.
  • Always check by substituting each solution back into the equation.
Practice quadratic formula problems Try the quadratic solver See it on FiboMap
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