# How do you solve 8x^2-5=-4x?

Jun 15, 2018

$\implies x \approx 0.57915$

or

$x \approx - 1.07915$

#### Explanation:

$8 {x}^{2} - 5 = - 4 x$

$\implies 8 {x}^{2} - 5 + 4 x = 0$

$\implies 8 {x}^{2} + 4 x - 5 = 0$ --------(1)

using quadratic equation formula:

$\implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute the corresponding values of a,b, and c from (1)

$\implies x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \times 8 \times \left(- 5\right)}}{2 \times 8}$

$\implies x = \frac{- 4 \pm \sqrt{16 + 160}}{16}$

$\implies x = \frac{- 4 \pm \sqrt{176}}{16}$

$\implies x \approx - \frac{4}{16} \pm \frac{13.266499}{16}$ ---truncating value of $\sqrt{176}$

$\implies x \approx - \frac{1}{4} \pm \frac{13.2665}{16}$

$\implies x \approx - 0.25 \pm 0.82915$

$\implies x \approx - 0.25 + 0.82915 \mathmr{and} x \approx - 0.25 - 0.82915$

$\implies x \approx 0.57915$

or

$x \approx - 1.07915$