Answer
$x=-2+\sqrt{2}i$ or $x=-2-\sqrt{2}i$
Work Step by Step
$ x^{2}+4x+6=0\qquad$... solve with the Quadractic formula. $a=1,\ b=4,\ c=6$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $4, \ a$ for $1$ and $c$ for $6$.
$ x=\displaystyle \frac{-4\pm\sqrt{(4)^{2}-4\cdot(6)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$x=\displaystyle \frac{-4\pm\sqrt{16-24}}{2}$
$x=\displaystyle \frac{-4\pm\sqrt{-8}}{2}$
$ x=\displaystyle \frac{-4+\sqrt{-1\cdot 4\cdot 2}}{}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$)
$x=\displaystyle \frac{-4\pm 2\sqrt{2}i}{2}$
$ x=-2\pm\sqrt{2}i\qquad$... the symbol $\pm$ indicates two solutions.
$x=-2+\sqrt{2}i$ or $x=-2-\sqrt{2}i$
