# How do you solve the quadratic with complex numbers given -x^2+4x-5=0?

Sep 10, 2016

Use the quadratic formula. $x = 2 \pm i$

#### Explanation:

$- {x}^{2} + 4 x - 5 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = - 1$
$b = 4$
$c = - 5$

$x = \frac{- 4 \pm \sqrt{{4}^{2} - \left(4 \cdot - 1 \cdot - 5\right)}}{2 \cdot - 1}$

$x = \frac{- 4 \pm \sqrt{16 - 20}}{-} 2$

$x = \frac{- 4 \pm \sqrt{- 4}}{-} 2$

$x = \frac{- 4 \pm \sqrt{4} \sqrt{- 1}}{-} 2$

$\sqrt{- 1} = i$

$x = \frac{- 4 \pm 2 i}{-} 2$

Divide both of the terms in the numerator by the denominator.

$x = \frac{- 4}{-} 2 \pm \frac{2 i}{-} 2$

$x = 2 \pm i$ which can also be written as $x = 2 + i$ and $x = 2 - i$