# How do you solve 2x ^ { 2} + x = 5?

Dec 26, 2016

-0.53, 1.35

#### Explanation:

Bring to standard form:
$y = 2 {x}^{2} + x - 5 = 0$
$D = {d}^{2} = 1 + 40 = 41 - \to d = \pm \sqrt{41}$
There are 2 real rots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{4} \pm \frac{\sqrt{41}}{4}$

$x 1 = \left(- 1 + 6.40\right) 4 = \frac{5.40}{4} = 1.35$
$x 2 = \frac{- 1 - 6.40}{14} = - \frac{7.40}{14} = - 0.53$