How to Solve Quadratic Equations: 3 Methods Explained Step by Step

Quadratic equations appear in almost every maths exam, from Matric and O Level to FSc, A Level and first-year university courses. The good news is that you only need three reliable methods to solve any of them. In this guide we walk through each method step by step, show you when to use which, and point out the mistakes that cost students marks.

What Is a Quadratic Equation?

A quadratic equation is any equation that can be written in the standard form

$$ ax^2 + bx + c = 0, \quad a \neq 0 $$

Here \(a\), \(b\) and \(c\) are real numbers. The highest power of \(x\) is 2, which is why a quadratic always has at most two solutions (also called roots).

Before using any method, always rearrange the equation so that one side equals zero. For example, \(x^2 = 5x - 6\) must first become \(x^2 - 5x + 6 = 0\).

Method 1: Factorization

Factorization is the fastest method when the quadratic splits neatly into two brackets.

Steps

1. Write the equation in standard form.
2. Find two numbers that multiply to \(ac\) and add to \(b\).
3. Split the middle term and factor by grouping.
4. Set each bracket equal to zero and solve.

Worked Example 1

Solve \(x^2 - 5x + 6 = 0\).

Here \(a = 1\), \(b = -5\), \(c = 6\). We need two numbers that multiply to \(6\) and add to \(-5\). These are \(-2\) and \(-3\).

$$ x^2 - 2x - 3x + 6 = 0 $$

$$ x(x - 2) - 3(x - 2) = 0 $$

$$ (x - 2)(x - 3) = 0 $$

So \(x - 2 = 0\) or \(x - 3 = 0\), giving \(x = 2\) or \(x = 3\).

Check: \(2^2 - 5(2) + 6 = 4 - 10 + 6 = 0\). Correct.

Factorization works best when the roots are whole numbers or simple fractions. If you cannot find suitable numbers within a minute, switch to another method.

Method 2: Completing the Square

Completing the square turns the left side into a perfect square. It always works, and it is the method used to derive the quadratic formula itself.

Steps

1. Make the coefficient of \(x^2\) equal to 1 (divide through by \(a\) if needed).
2. Move the constant term to the right side.
3. Add \(\left(\frac{b}{2}\right)^2\) to both sides.
4. Write the left side as a squared bracket.
5. Take the square root of both sides (remember \(\pm\)) and solve.

Worked Example 2

Solve \(x^2 + 6x - 7 = 0\).

Move the constant:

$$ x^2 + 6x = 7 $$

Half of 6 is 3, and \(3^2 = 9\). Add 9 to both sides:

$$ x^2 + 6x + 9 = 16 $$

$$ (x + 3)^2 = 16 $$

Take square roots:

$$ x + 3 = \pm 4 $$

So \(x = -3 + 4 = 1\) or \(x = -3 - 4 = -7\).

Check: \(1^2 + 6(1) - 7 = 0\) and \((-7)^2 + 6(-7) - 7 = 49 - 42 - 7 = 0\). Both correct.

Method 3: The Quadratic Formula

The quadratic formula solves every quadratic equation, including those with irrational or complex roots:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

The expression \(b^2 - 4ac\) is called the discriminant, written \(D\). It tells you the nature of the roots:

• \(D > 0\): two distinct real roots
• \(D = 0\): two equal (repeated) real roots
• \(D < 0\): no real roots (two complex roots)

Worked Example 3

Solve \(2x^2 + 3x - 2 = 0\).

Identify \(a = 2\), \(b = 3\), \(c = -2\). Calculate the discriminant:

$$ D = 3^2 - 4(2)(-2) = 9 + 16 = 25 $$

Substitute into the formula:

$$ x = \frac{-3 \pm \sqrt{25}}{2(2)} = \frac{-3 \pm 5}{4} $$

So \(x = \frac{-3 + 5}{4} = \frac{1}{2}\) or \(x = \frac{-3 - 5}{4} = -2\).

Worked Example 4 (Irrational Roots)

Solve \(3x^2 - 4x - 2 = 0\).

Here \(a = 3\), \(b = -4\), \(c = -2\).

$$ D = (-4)^2 - 4(3)(-2) = 16 + 24 = 40 $$

$$ x = \frac{4 \pm \sqrt{40}}{6} = \frac{4 \pm 2\sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3} $$

Numerically, \(x \approx 1.721\) or \(x \approx -0.387\). This equation cannot be factorized nicely, which shows why the formula is so valuable.

Which Method Should You Use?

• Factorization: when the numbers are small and the roots look like integers.
• Completing the square: when the question asks for it, or when you need the vertex form of a parabola.
• Quadratic formula: whenever factorization fails, or when the roots are irrational or complex.

A quick tip: calculate the discriminant first. If it is a perfect square (like 1, 4, 9, 25), the quadratic factorizes. If not, go straight to the formula.

Common Mistakes

• Not setting the equation to zero. Solving \(x^2 = 4x\) by dividing by \(x\) loses the root \(x = 0\). Instead write \(x^2 - 4x = 0\), so \(x(x - 4) = 0\).
• Forgetting the \(\pm\). Square roots give two values. Missing the negative root halves your answer.
• Sign errors with \(b\). If \(b = -4\), then \(-b = 4\) and \(b^2 = 16\), not \(-16\).
• Dividing only part of the numerator. In the formula, \(2a\) divides the whole expression \(-b \pm \sqrt{D}\), not just \(\sqrt{D}\).
• Skipping the check. Substituting your roots back takes seconds and catches most errors.

Key Takeaways

• Always write the quadratic in the form \(ax^2 + bx + c = 0\) before solving.
• Factorization is quickest when the discriminant is a perfect square.
• Completing the square always works and leads to the vertex form.
• The quadratic formula solves every quadratic; the discriminant reveals the nature of the roots.
• Check your answers by substitution to secure full marks.

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