# How do you solve 2x^2-x=-5 by completing the square?

Jan 12, 2018

This question doesn't have any real solutions. However it does have two complex solutions;
$\textcolor{red}{x = \frac{1 \pm i \sqrt{39}}{4}}$

#### Explanation:

$2 {x}^{2} - x = - 5$
$2 {x}^{2} - x + 5 = 0$
$2 \left({x}^{2} - \frac{x}{2} + \frac{5}{2}\right) = 0$
${x}^{2} - \frac{x}{2} + \frac{5}{2} = 0$
${\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16} + \frac{5}{2} = 0$
${\left(x - \frac{1}{4}\right)}^{2} + \frac{39}{16} = 0$
${\left(x - \frac{1}{4}\right)}^{2} = - \frac{39}{16}$
$x - \frac{1}{4} = \pm \sqrt{- \frac{39}{16}}$
$x - \frac{1}{4} = \pm \frac{\sqrt{- 39}}{4} = \pm \frac{\sqrt{39} \times \sqrt{- 1}}{4}$
$x = \frac{1}{4} \pm \frac{\sqrt{39} \times \sqrt{- 1}}{4} = \frac{1}{4} \pm \frac{\sqrt{39} \times i}{4}$
$\textcolor{red}{x = \frac{1 \pm i \sqrt{39}}{4}}$

I hope it helps :)