# How do you solve -5x^2-2x-4 = 0 ?

Nov 11, 2017

$x = \frac{1}{5} \left(- 1 \pm \sqrt{19} i\right)$

#### Explanation:

Complete the square, then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(5 x + 1\right)$ and $b = \sqrt{19} i$ as follows:

$0 = - 5 \left(- 5 {x}^{2} - 2 x - 4\right)$

$\textcolor{w h i t e}{0} = 25 {x}^{2} + 10 x + 20$

$\textcolor{w h i t e}{0} = {\left(5 x\right)}^{2} + 2 \left(5 x\right) + 1 + 19$

$\textcolor{w h i t e}{0} = {\left(5 x + 1\right)}^{2} + {\left(\sqrt{19}\right)}^{2}$

$\textcolor{w h i t e}{0} = {\left(5 x + 1\right)}^{2} - {\left(\sqrt{19} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(5 x + 1\right) - \sqrt{19} i\right) \left(\left(5 x + 1\right) + \sqrt{19} i\right)$

$\textcolor{w h i t e}{0} = \left(5 x + 1 - \sqrt{19} i\right) \left(5 x + 1 + \sqrt{19} i\right)$

Hence:

$5 x = - 1 \pm \sqrt{19} i$

So:

$x = \frac{1}{5} \left(- 1 \pm \sqrt{19} i\right)$