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

Jan 30, 2018

$x = 1.2 - 0.1 \sqrt{26} \textcolor{red}{i} , 1.2 + 0.1 \sqrt{26} \textcolor{red}{i}$

#### Explanation:

$- 5 {x}^{2} + 12 x - 8 = 0$

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

$a = - 5 , b = 12 , c = - 8$

$x = \frac{- 12 \pm \sqrt{{12}^{2} - \left(4 \cdot - 5 \cdot - 8\right)}}{2 \cdot - 5}$

$x = \frac{- 12 \pm \sqrt{144 - 160}}{- 10}$

$x = 1.2 \pm 0.1 \sqrt{- 26}$

$x = 1.2 \pm 0.1 \sqrt{26} \textcolor{red}{i}$