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Quadratic Formula (Cambridge (CIE) IGCSE Additional Maths): Revision Note

Exam code: 0606

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Updated on 2 July 2026

Quadratic formula

How do I use the quadratic formula to solve a quadratic equation?

A quadratic equation has the form:
ax² + bx + c = 0 (as long as a ≠ 0)
- you need "= 0" on one side
The quadratic formula is a formula that gives both solutions:
- $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Read off the values of a, b and c from the equation
Substitute these into the formula
- write this line of working in the exam
- Put brackets around any negative numbers being substituted in
To solve $2 x^{2} - 5 x + 2 = 0$ using the quadratic formula:
- a = 2, b = -5 and c = 2
- $x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 \times 2 \times 2}}{2 \times 2}$
- Carefully simplify by doing the calculation in parts $x = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4}$
- Separate the + and - to get the two answers $\begin{array}{rcl} x & = & \frac{5 + 3}{4} = \frac{8}{4} = 2 \\ x & = & \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2} \end{array}$
On the non-calculator paper, answers may be required in exact form
On the calculator paper, you might have to round your answers
- Give your answers in exact form before rounding in case you round incorrectly

Examiner Tips and Tricks

On the calculator paper you will be able to check your solutions using the quadratic equation solver feature if your calculator has one
Always look for how the question wants you to leave your final answers
- for example, correct to 2 decimal places

Worked Example

Use the quadratic formula, without a calculator, to find the exact solutions of the equation $12 x^{2} - 17 x + 6 = 0$ .

Write down the values of a, b and c

$a = 12, b = - 17, c = 6$

Substitute these values into the quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Put brackets around any negative numbers

$x = \frac{- (- 17) \pm \sqrt{{(- 17)}^{2} - 4 \times 12 \times 6}}{2 \times 12}$

Simplify

$\begin{array}{rcl} x & = & \frac{17 \pm \sqrt{289 - 288}}{24} \\ = & \frac{17 \pm \sqrt{1}}{24} \end{array}$

Work out the + and - parts separately to get the two solutions

$x = \frac{17 + 1}{24} = \frac{18}{24} x = \frac{17 - 1}{24} = \frac{16}{24}$

$x = \frac{3}{4}, x = \frac{2}{3}$

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