# How do you solve by completing the square: 4x^2 + 8x + 20 = 0?

Apr 2, 2015

Try this:
$4 {x}^{2} + 8 x = - 20$
${x}^{2} + \frac{8}{4} x = - \frac{20}{4}$
${x}^{2} + 2 x = - 5$
${x}^{2} + 2 x + 1 - 1 = - 5$
${\left(x + 1\right)}^{2} = - 5 + 1$
${\left(x + 1\right)}^{2} = - 4$
$x + 1 = \pm \sqrt{- 4}$ NEGATIVE ROOT!
If you already saw complex numbers you have:
$x + 1 = \pm \left[\sqrt{4} \sqrt{- 1}\right]$
$x = - 1 \pm 2 i$