# How do you solve the quadratic with complex numbers given 5x^2+12x+8=0?

Dec 25, 2016

The solutions are $S = \left\{- \frac{6}{5} - \frac{2}{5} i , - \frac{6}{5} + \frac{2}{5} i\right\}$

#### Explanation:

We need $a {x}^{2} + b x + c = 0$

$\Delta = {b}^{2} - 4 a c$

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

The equation is $5 {x}^{2} + 12 x + 8 = 0$

$\Delta = {12}^{2} - 4 \cdot 5 \cdot 8 = 144 - 160 = - 16$

As, $\Delta < 0$, there are no roots in $\mathbb{R}$ but in $\mathbb{C}$

$x = \frac{- 12 \pm 4 i}{10}$

The roots are ${x}_{1} = - \frac{6}{5} - \frac{2}{5} i$ and ${x}_{2} = - \frac{6}{5} + \frac{2}{5} i$