# How do you describe the nature of the roots of the equation x^2+4x=96?

Sep 16, 2016

8 and -12

#### Explanation:

Bring equation to standard form:
$y = {x}^{2} + 4 x - 96 = 0$
$D = {b}^{2} - 4 a c = 16 + 384 = 400 > 0$ --> $d = \pm 20$
Since D > 0, there are 2 real roots. The roots have opposite signs (ac > 0).
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{4}{2} \pm \frac{20}{2} = - 2 \pm 10.$
x1 = -2 + 10 = 8
x2 = -2 - 10 = -12