Answer
$x=-2+i $ is a solution.
Work Step by Step
$ x^{2}+4x+5=0\qquad$...apply the quadratic formula $x_{1,2}=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a},\ a=1,b=4,c=5$
$ x=\displaystyle \frac{-4\pm\sqrt{4^{2}-4\cdot 1\cdot 5}}{2\cdot 1}\qquad$...We take "$+$" to find one of the solutions...
$ x=\displaystyle \frac{-4+\sqrt{4^{2}-4\cdot 1\cdot 5}}{2\cdot 1}\qquad$...Simplify.
$x=\displaystyle \frac{-4+\sqrt{16-4\cdot 1\cdot 5}}{2}$
$x=\displaystyle \frac{-4+\sqrt{16-20}}{2}$
$ x=\displaystyle \frac{-4+\sqrt{-4}}{2}\qquad$...Apply radical rule: $\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}$
$ x=\displaystyle \frac{-4+\sqrt{-1}\sqrt{4}}{2}\qquad$...Apply imaginary number rule: $\sqrt{-1}=i$
$ x=\displaystyle \frac{-4+i\sqrt{4}}{2}\qquad$...Factor $-4+i\sqrt{4}$
$ x=\displaystyle \frac{2(-2+i)}{2}\qquad$...Simplify.
$x=-2+i $
