In this article
A quadratic equation is any equation that can be written as ax² + bx + c = 0, with a ≠ 0. The x² term is what makes it quadratic. These equations describe thrown balls, profit curves, bridge cables and the area of fields, which is why every maths course teaches them.
The formula step by step
Start with x² − 5x + 6 = 0. Here a = 1, b = −5 and c = 6. Substitute into the formula and work from the inside out.
- 1Identify a, b, ca = 1, b = −5, c = 6.
- 2DiscriminantD = (−5)² − 4 × 1 × 6 = 25 − 24 = **1**.
- 3Square root√1 = 1.
- 4Two rootsx = (5 + 1) ÷ 2 = **3** and x = (5 − 1) ÷ 2 = **2**.
- 5Check3² − 15 + 6 = 0 and 2² − 10 + 6 = 0. Both work.
Note that −b is −(−5) = +5. Dropping that sign is the most common slip.
The discriminant tells you what to expect
The part under the square root, D = b² − 4ac, is checked first. You cannot take the square root of a negative number and stay with real numbers, so the sign of D splits quadratics into three cases.
When D < 0 the two roots are complex numbers: x² + 2x + 5 = 0 gives x = −1 ± 2i.
| Equation | a, b, c | D | Roots |
|---|---|---|---|
| x² − 5x + 6 = 0 | 1, −5, 6 | 1 | 2 and 3 |
| 2x² + 3x − 2 = 0 | 2, 3, −2 | 25 | 0.5 and −2 |
| x² − 6x + 9 = 0 | 1, −6, 9 | 0 | 3 (repeated) |
| x² + 2x + 5 = 0 | 1, 2, 5 | −16 | −1 + 2i and −1 − 2i |
For 2x² + 3x − 2: x = (−3 ± 5) ÷ 4, so 2 ÷ 4 = 0.5 and −8 ÷ 4 = −2.
Seeing the roots on a graph
Plot y = x² − 5x + 6 and the curve is a U-shaped parabola. It crosses zero exactly at the two roots. The lowest point is halfway between them, at x = 5 ÷ 2 = 2.5, where y = −0.25.
| x | 0 | 1 | 2 | 2.5 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|
| y = x² − 5x + 6 | 6 | 2 | 0 | −0.25 | 0 | 2 | 6 |
The curve is symmetric about the vertex at x = 2.5. Zero appears at x = 2 and x = 3, the two roots, at equal distances either side of the lowest point.
When a is positive the parabola opens upward and the vertex is a minimum. When a is negative it opens downward and the vertex is a maximum, as in a thrown ball.
Three ways to solve one
| Method | Best when | Weakness |
|---|---|---|
| Factoring | The numbers are small and the roots are whole or simple fractions | Hard to spot when roots are messy |
| Quadratic formula | Always works, for any a, b and c | More arithmetic, easy to slip on signs |
| Completing the square | You need the vertex form or want to see where the formula comes from | Slower for one-off answers |
Factoring example: x² − 5x + 6 = (x − 2)(x − 3). Each bracket is zero at a root.
Completing the square rewrites x² − 5x + 6 as (x − 2.5)² − 0.25. Setting that to zero gives (x − 2.5)² = 0.25, so x − 2.5 = ±0.5 and x = 2 or 3. The quadratic formula is exactly this process done once, in general.
A quick check using sum and product
If the roots are r₁ and r₂, then r₁ + r₂ = −b ÷ a and r₁ × r₂ = c ÷ a. It is a fast way to confirm your answers.
For 2x² + 3x − 2 the roots 0.5 and −2 sum to −1.5 = −3 ÷ 2 and multiply to −1 = −2 ÷ 2. Both checks pass.
A real case: a ball thrown upward
A ball is thrown straight up at 20 m/s from a height of 1.5 m. Ignoring air resistance, its height after t seconds is h = −4.9t² + 20t + 1.5 metres. When does it hit the ground? Set h = 0 and solve.
- 1Identifya = −4.9, b = 20, c = 1.5.
- 2DiscriminantD = 20² − 4(−4.9)(1.5) = 400 + 29.4 = **429.4**, so √D ≈ 20.72.
- 3Two rootst = (−20 ± 20.72) ÷ (−9.8), giving **t ≈ 4.16 s** and t ≈ −0.07 s.
- 4Choose the sensible oneTime cannot be negative here, so the ball lands after about **4.16 seconds**.
The negative root is a real solution of the equation but not of the situation. Always ask which answers make sense.
The vertex gives the peak: at t = 2.04 s, h = 1.5 + 20² ÷ (4 × 4.9) ≈ 21.9 m.
It rises for about 2 seconds, peaks near 21.9 m and then falls back to the ground. The path is a parabola.
Common mistakes
- Not rearranging to zero first. The equation must be in the form ax² + bx + c = 0 before you read off a, b and c.
- Sign errors on b. If b = −5, then −b = +5 and b² = 25.
- Dividing only part of the numerator by 2a. The whole of (−b ± √D) is divided by 2a.
- Forgetting a = 0 case. If a = 0 the equation is linear, not quadratic.
- Keeping answers that do not fit the problem, such as negative times or lengths.
Questions people ask
What is the quadratic formula?
For ax² + bx + c = 0 the solutions are x = (−b ± √(b² − 4ac)) ÷ 2a. It works for every quadratic equation.
What is the discriminant?
It is b² − 4ac, the expression under the square root. If it is positive there are two real roots, if zero there is one repeated root, and if negative there are no real roots.
Can a quadratic equation have no solution?
It has no real solution when the discriminant is negative. It always has two solutions among the complex numbers, though.
How do I find the vertex of a parabola?
The x-coordinate of the vertex is −b ÷ 2a. Substitute it back into the equation to find the y-coordinate.
How do I check my roots?
Substitute each root into the original equation, or use the sum and product: the roots add to −b/a and multiply to c/a.
When should I use factoring instead of the formula?
Factoring is quicker when the coefficients are small and the roots are whole numbers. The formula is the reliable choice when they are not.