18 (a) Solve by factorisation - OCR - GCSE Maths - Question 18 - 2017 - Paper 1

To solve this quadratic equation, we will use the quadratic formula:

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

Here, a = 3, b = 2, and c = -3. First, calculate the discriminant:

$b^2 - 4ac = 2^2 - 4(3)(-3) = 4 + 36 = 40$

Now, substituting the values into the quadratic formula:

$x = \frac{-2 \pm \sqrt{40}}{2(3)} = \frac{-2 \pm 2\sqrt{10}}{6} = \frac{-1 \pm \frac{\sqrt{10}}{3}}{3}$

Calculating both values:

1. Using the + sign: $x_1 = \frac{-1 + \frac{2\sqrt{10}}{6}}{3} \approx 0.72$
2. Using the - sign: $x_2 = \frac{-1 - \frac{2\sqrt{10}}{6}}{3} \approx -1.39$

So, the values are:

x = 0.72 or x = -1.39.