How to Solve Quadratic Equations: 3 Methods With Worked Examples

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

ax² + bx + c = 0, where a ≠ 0

It describes a parabola, the U-shaped curve of a thrown ball, a satellite dish or a profit curve. “Solving” means finding the values of x where the parabola crosses the x-axis: the roots. There are three standard methods.

Method 1: Factoring

If the quadratic splits neatly into two brackets, factoring is the fastest route.

Solve x² − 5x + 6 = 0.

Find two numbers that multiply to +6 and add to −5: that’s −2 and −3.

(x − 2)(x − 3) = 0

If a product is zero, one factor must be zero, so x = 2 or x = 3.

Factoring works well when a = 1 and the roots are whole numbers. When it doesn’t come out cleanly within a few seconds, switch methods.

Method 2: Completing the square

This method rewrites the equation as a perfect square and always works.

Solve x² + 6x − 7 = 0.

Move the constant: x² + 6x = 7
Halve the x-coefficient (6 → 3), square it (9) and add to both sides: x² + 6x + 9 = 16
Write as a square: (x + 3)² = 16
Square-root both sides: x + 3 = ±4
Solve: x = 1 or x = −7

Completing the square also reveals the vertex form a(x − h)² + k, which tells you the parabola’s turning point (h, k) directly.

Method 3: The quadratic formula

The universal method, derived by completing the square on the general equation:

x = (−b ± √(b² − 4ac)) ÷ 2a

Solve 2x² + 3x − 2 = 0 (a = 2, b = 3, c = −2):

b² − 4ac = 9 − 4(2)(−2) = 9 + 16 = 25
√25 = 5
x = (−3 ± 5) ÷ 4 → x = 0.5 or x = −2

MathKit’s Quadratic tab applies this formula instantly and shows every intermediate value.

The discriminant: what kind of roots?

The expression under the square root, D = b² − 4ac, is the discriminant. It tells you what to expect before you solve:

Discriminant	Roots	Parabola
D > 0	Two different real roots	Crosses the x-axis twice
D = 0	One repeated real root	Touches the x-axis once
D negative	Two complex roots	Never touches the x-axis

Complex roots example

x² + 2x + 5 = 0: D = 4 − 20 = −16. The square root of −16 is 4i, so:

x = (−2 ± 4i) ÷ 2 = −1 ± 2i

There are no real solutions; the parabola sits entirely above the x-axis.

Finding the vertex

The vertex (turning point) of y = ax² + bx + c is at:

x = −b ÷ 2a
y = c − b² ÷ 4a

For x² − 5x + 6, the vertex is at (2.5, −0.25). If a is positive the vertex is a minimum; if a is negative, a maximum. This is how you find the peak height of a projectile or the price that maximises profit.

Which method should I use?

Try factoring for a few seconds if a = 1 and the numbers are small.
Use the formula for everything else; it never fails.
Complete the square when you need vertex form or are proving something.

Check your answer

Substitute each root back into the original equation. For x = 3 in x² − 5x + 6: 9 − 15 + 6 = 0 ✓.

Need to factor the constant term quickly? Use the prime factorisation tool. More in GCD and LCM Explained: How to Find the Greatest Common Divisor and Least Common Multiple.
Analysing data instead of equations? See Mean, Median, Mode and Standard Deviation: A Practical Guide.

Try the examples above in MathKit below. Open the Quadratic tab and enter a, b and c.