# How do you solve by completing the square 3x^2 + 6x +12 = 0?

Mar 30, 2015

First divide both sides of the equation by the coefficient of ${x}^{2}$ (in this case $3$)
${x}^{2} + 2 x + 4 = 0$

If ${x}^{2} + 2 x$ are the first 2 terms of an expression from
${\left(x + a\right)}^{2}$
then $a = 1$

Rewrite the left-side of the expression as
$\left({x}^{2} + 2 x + 1\right) + \left(3\right) = 0$

${\left(x + 1\right)}^{2} = - 3$

This can not be solved with any real number value for $x$
as can be seen from the graph of the original equation below (it never is equal to $0$)
graph{3x^2+6x+4 [-16.03, 16, -1.31, 14.71]}

However, if we allow complex solutions
$x + 1 = \pm \sqrt{- 3}$

$x = - 1 - \sqrt{3} i$ and $x = - 1 + \sqrt{3} i$