# How do you solve 2x^2 -3x - 7 = 0 using the quadratic formula?

Mar 25, 2016

$x = - \frac{1}{2} + \sqrt{5}$ or $x = - \frac{1}{2} - \sqrt{5}$

#### Explanation:

$2 {x}^{2} - 3 x - 7 = 0$

Identify the a, b, and c terms

$a {x}^{2} - b x - c = 0$
$a = 2$
$b = - 3$
$c = - 7$

$x = \frac{- b \left(\frac{+}{-}\right) \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \left(- 2 \left(\frac{+}{-}\right) \frac{\sqrt{{2}^{2} - 4 \left(- 3\right) \left(- 7\right)}}{2 \left(2\right)}\right)$

$x = \frac{- 2 \left(\frac{+}{-}\right) \sqrt{4 - 84}}{4}$

$x = \frac{- 2 \left(\frac{+}{-}\right) \sqrt{- 80}}{4}$ simplify the radical $i = \sqrt{- 1}$

$x = \frac{- 2 \left(\frac{+}{-}\right) 4 i \sqrt{5}}{4}$ simplify the fractions

$x = - \frac{1}{2} + \sqrt{5}$ or $x = - \frac{1}{2} - \sqrt{5}$