Answer
$x=-2\pm\sqrt{2}i$
Work Step by Step
$x^{2}+4x+6=0$
Take the $6$ to the right side of the equation:
$x^{2}+4x=-6$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular case, $b=4$
$x^{2}+4x+\Big(\dfrac{4}{2}\Big)^{2}=-6+\Big(\dfrac{4}{2}\Big)^{2}$
$x^{2}+4x+4=-6+4$
$x^{2}+4x+4=-2$
Factor the left side of the equation, which is a perfect square trinomial:
$(x+2)^{2}=-2$
Take the square root of both sides of the equation:
$\sqrt{(x+2)^{2}}=\sqrt{-2}$
$x+2=\pm\sqrt{2}i$
Solve for $x$:
$x=-2\pm\sqrt{2}i$
