# How do you solve sqrt(6x^2 + 5x) = 2?

Jul 29, 2018

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

#### Explanation:

$\sqrt{6 {x}^{2} + 5 x} = 2$

Square both sides
$6 {x}^{2} + 5 x = 4$

$6 {x}^{2} + 5 x - 4 = 0$

$x = \frac{- 5 \pm \sqrt{25 + 96}}{12}$
$x = \frac{- 5 \pm \sqrt{121}}{12}$
$x = \frac{- 5 \pm 11}{12}$
$x = \frac{- 5 + 11}{12}$ or $x = \frac{- 5 - 11}{12}$
$x = \frac{1}{2}$ or $x = - \frac{4}{3}$
$6 {x}^{2} + 5 x - 4 = \left(x - \frac{1}{2}\right) \left(x + \frac{4}{3}\right) = \left(2 x - 1\right) \left(3 x + 4\right)$