# How do you solve x^2+20x+104=0 by completing the square?

Aug 13, 2016

$x = - 10 \pm 2 i$

#### Explanation:

Move the constant term to the RHS.

${x}^{2} + 20 x = - 104$

Add the square of half the coefficient of the $x$ term to both sides:

${x}^{2} + 20 x + \textcolor{red}{{10}^{2}} = - 104 + \textcolor{red}{{10}^{2}}$

This becomes:

${\left(x + 10\right)}^{2} = - 104 + 100$

${\left(x + 10\right)}^{2} = - 4$

Take square roots of both sides.

$x + 10 = \pm \sqrt{- 4} = \pm \sqrt{4 {i}^{2}} = \pm 2 i$

$x = - 10 \pm 2 i$