Answer
$-2-i$ satisfies the equation.
Work Step by Step
$ x^{2}+4x+5=0\qquad$...substitute $x$ for $-2-i$.
$(-2-i)^{2}+4(-2-i)+5=0\qquad$...Apply the Perfect Square Formula: $(a-b)^{2}=a^{2}-2ab+b^{2},\ a=-2,\ b=i$
$(-2)^{2}-2(-2)i+i^{2}+4(-2-i)+5=0$
$ 4+2\cdot 2i+i^{2}+4(-2-i)+5=0\qquad$...Apply imaginary number rule: $\quad i^{2}=-1$
$ 4+4i-1+4(-2-i)+5=0\qquad$...Simplify.
$ 8+4i+4(-2-i)=0\qquad$...Apply the distributive property: $\quad a(b-c)=ab-ac$
$ 8+4i+4(-2)-4i=0\qquad$...Simplify.
$8-8+4i-4i=0$
$0=0$
