A quadratic equation is anything that can be rearranged into ax² + bx + c = 0 with a ≠ 0. It has at most two solutions, called roots. The first job is always the same: get everything onto one side, equal to zero, with the x² term first.
Faced with x² = 5x, it's tempting to divide both sides by x and get x = 5. That throws away the other root, x = 0. Rearrange to x² − 5x = 0, factorise to x(x − 5) = 0, and both appear.
Method 1: factorising
Best when the numbers are small whole numbers and the equation factorises "nicely". It's the fastest method when it works.
Solve x² − 5x + 6 = 0.
Find two numbers that multiply to c = 6 and add to b = −5. Trying pairs that multiply to 6: (1, 6), (2, 3), (−1, −6), (−2, −3). Only −2 and −3 add to −5.
(x − 2)(x − 3) = 0 so x = 2 or x = 3
If two things multiply to zero, one of them must be zero. That's the whole idea.
When a isn't 1, look for factors of a × c instead. For 2x² − 7x + 3 = 0, a × c = 6, and the pair that adds to −7 is −1 and −6. Split the middle term and group:
2x² − x − 6x + 3 = 0
x(2x − 1) − 3(2x − 1) = 0
(2x − 1)(x − 3) = 0 so x = ½ or x = 3
The 10-second test: will it factorise?
Work out the discriminant b² − 4ac. With whole-number coefficients, the quadratic factorises into whole-number brackets only if the discriminant is a perfect square (0, 1, 4, 9, 16, …). For 2x² − 7x + 3, that's 49 − 24 = 25 = 5², so yes. For x² + 4x − 3, it's 16 + 12 = 28, not a square, so don't waste time hunting for factors: go straight to the formula.
Method 2: completing the square
Best when a = 1 and b is even, and essential when you also want the vertex (the turning point) of the graph.
Solve x² + 6x − 7 = 0.
Half of b is 3. Since (x + 3)² = x² + 6x + 9, rewrite x² + 6x as (x + 3)² − 9:
(x + 3)² − 9 − 7 = 0
(x + 3)² = 16
x + 3 = ±4
x = 1 or x = −7
As a bonus, (x + 3)² − 16 tells you the vertex of y = x² + 6x − 7 is at (−3, −16).
If a isn't 1, divide through by a first. Completing the square on the general equation is exactly how the quadratic formula is derived, which is shown on the quadratic solver page.
Method 3: the quadratic formula
x = ( −b ± √(b² − 4ac) ) / 2a
It always works, which makes it the right choice for awkward coefficients, decimals, or when the discriminant isn't a perfect square.
Solve 3x² + 4x − 2 = 0. Here a = 3, b = 4, c = −2.
b² − 4ac = 16 − 4(3)(−2) = 16 + 24 = 40
x = (−4 ± √40) / 6
x = (−4 ± 6.3246) / 6
x = 0.3874 or x = −1.7208
In exact form, √40 = 2√10, so x = (−2 ± √10) / 3.
Check it with full working.
Open in the solverWhat the discriminant tells you
| b² − 4ac | Roots | Graph |
|---|---|---|
| Positive | Two different real roots | Crosses the x-axis twice |
| Zero | One repeated root, x = −b/2a | Touches the x-axis at the vertex |
| Negative | No real roots (two complex roots) | Never meets the x-axis |
Exam questions often ask "for what values of k does this have two real roots?" That's just the inequality b² − 4ac > 0 with k inside a, b or c.
Checking your answer in 20 seconds
For ax² + bx + c = 0, the two roots always satisfy:
x₁ + x₂ = −b / a x₁ × x₂ = c / a
For 3x² + 4x − 2: the sum should be −4/3 = −1.3333 and the product −2/3 = −0.6667. Our roots: 0.3874 + (−1.7208) = −1.3334 and 0.3874 × (−1.7208) = −0.6666. Both match to within rounding of the last digit.
The other check is to substitute a root back into the original equation; it should come out as (very nearly) zero.
Which method should you use?
- Small whole numbers, perfect-square discriminant → factorise.
- a = 1, even b, or you need the vertex → complete the square.
- Anything else → the formula.
- No c term (like 4x² − 12x = 0) → take out the common factor: 4x(x − 3) = 0.
- No b term (like x² − 20 = 0) → rearrange and square-root: x = ±√20 = ±4.472. Don't forget the ±.
A real problem
A rectangular garden has a perimeter of 30 m and an area of 50 m². What are its sides?
If one side is x, the other is 15 − x (half the perimeter minus x). So x(15 − x) = 50, which rearranges to x² − 15x + 50 = 0. Numbers that multiply to 50 and add to −15: −5 and −10. So the garden is 5 m by 10 m. Both roots give the same rectangle, just turned round.