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Solving quadratic equations - AQAUsing the quadratic formula - Higher

Solve quadratic equations by factorising, using formulae and completing the square. Each method also provides information about the corresponding quadratic graph.

Part of MathsAlgebra

Solving quadratic equations using the quadratic formula - Higher

The quadratic formula can be used to solve quadratic equations that will not factorise. You will need to learn this formula, as well as understanding how to use it.

Example

Solve \(x^2 + 6x - 12 = 0\) using the quadratic formula.

First, identify the value of \(a\), \(b\) and \(c\). In this example, \(a = 1\), \(b = 6\) and \(c = -12\).

Substitute these values into the quadratic formula:

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

\(x = \frac{(- 6) \pm \sqrt{6^2 - 4 \times 1 \times (-12)}}{2 \times 1}\)

\(x = \frac{(-6) \pm \sqrt{36 + 48}}{2}\)

\( x = \frac{(-6) \pm \sqrt{84}}{2}\)

To calculate the decimal answers, work out each answer in turn on a calculator.

\(x = \frac{(-6) \pm \sqrt{84}}{2}\)

The first solution is \(x = \frac{(-6) + \sqrt{84}}{2} = 1.58\) (2 dp).

The second solution is \(x = \frac{(-6) - \sqrt{84}}{2} = -7.58\) (2 dp).

Question

Solve \(2x^2 - 10x + 3 = 0\) using the quadratic formula.