# How do you solve the equation 2x^2-3x-3=0 by completing the square?

##### 1 Answer
Nov 10, 2016

$x = \frac{3 \pm \sqrt{15}}{2}$

#### Explanation:

$2 {x}^{2} - 3 x - 3 = 0$

divide by $2$ to make coefficient ${x}^{2} = 1$

${x}^{2} - \frac{3 x}{2} - \frac{3}{2} = 0$

$\left({x}^{2} - \frac{3 x}{2}\right) - \frac{3}{2} = 0$

square half coefficient of $x$ then add and subtract

$\left({x}^{2} - \frac{3 x}{2} + {\left(\frac{3}{4}\right)}^{2}\right) - \frac{3}{2} - {\left(\frac{3}{4}\right)}^{2} = 0$

the first bracket is now a perfect square

${\left(x - \frac{3}{2}\right)}^{2} - \frac{3}{2} - \frac{9}{4} = 0$

${\left(x - \frac{3}{2}\right)}^{2} - \frac{15}{4} = 0$

now rearrange for $x$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{15}{4}$

$x - \frac{3}{2} = \pm \sqrt{\frac{15}{4}}$

$x - \frac{3}{2} = \pm \frac{\sqrt{15}}{2}$

$x = \frac{3}{2} \pm \frac{\sqrt{15}}{2}$

$x = \frac{3 \pm \sqrt{15}}{2}$