# How do you solve  8x^2 + 2x - 3 = 0 using the quadratic formula?

May 4, 2016

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

#### Explanation:

Given:

$8 {x}^{2} + 2 x - 3 = 0$

This is in the form $a {x}^{2} + b x + c = 0$ with $a = 8$, $b = 2$ and $c = - 3$.

It has roots given by the quadratic formula:

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

$= \frac{- 2 \pm \sqrt{{2}^{2} - \left(4 \cdot 8 \cdot \left(- 3\right)\right)}}{2 \cdot 8}$

$= \frac{- 2 \pm \sqrt{4 + 96}}{16}$

$= \frac{- 2 \pm \sqrt{100}}{16}$

$= \frac{- 2 \pm 10}{16}$

That is:

$x = \frac{- 2 + 10}{16} = \frac{8}{16} = \frac{1}{2}$

or:

$x = \frac{- 2 - 10}{16} = \frac{- 12}{16} = - \frac{3}{4}$