# How do you solve x^2 + 3x = -4 ?

Apr 29, 2016

#### Answer:

There are no real roots.

#### Explanation:

$y = {x}^{2} + 3 x + 4 = 0$.
$D = {d}^{2} = {b}^{2} - 4 a c = 9 - 16 = - 7$
Since D < 0, there are no real roots. There are 2 complex roots.
${d}^{2} = - 7 = 7 {i}^{2}$ --> $d = \pm i \sqrt{7}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{3}{2} \pm i \frac{\sqrt{7}}{2} = \frac{- 3 \pm i \sqrt{7}}{2}$