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

Apr 23, 2016

$x = \frac{1}{2} + \frac{\sqrt{7}}{2}$ and
$x = \frac{1}{2} - \frac{\sqrt{7}}{2}$

#### Explanation:

In $2 {x}^{2} - 2 x - 3 = 0$, let us first divide this by $2$ and we get ${x}^{2} - x - \frac{3}{2} = 0$. As the coefficient of $x$ is $- 1$, let us add and subtract ${\left(- 1\right)}^{2} / 4$ or $\frac{1}{4}$, which gives us

${x}^{2} - x + \frac{1}{4} - \frac{3}{2} - \frac{1}{4} = 0$ or

${\left(x - \frac{1}{2}\right)}^{2} - \frac{7}{4} = 0$ or

${\left(x - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{7}}{2}\right)}^{2} = 0$ or

$\left(x - \frac{1}{2} - \frac{\sqrt{7}}{2}\right) \left(x - \frac{1}{2} + \frac{\sqrt{7}}{2}\right) = 0$ or

$x = \frac{1}{2} \pm \frac{\sqrt{7}}{2}$