Answer
$-3-4i$ satisfies the equation.
Work Step by Step
$ x^{2}+6x+25=0\qquad$...substitute $x$ for $-3-4i$.
$(-3-4i)^{2}+6(-3-4i)+25=0\qquad$...Apply the Perfect Square Formula: $(a+b)^{2}=a^{2}+2ab+b^{2},\ a=-3,\ b=-4i$
$(-3)^{2}+2(-3)\cdot(-4i)+(-4i)^{2}+6(-3-4i)+25=0$
$ 9+24i+16i^{2}+6(-3-4i)+25=0\qquad$...Apply imaginary number rule: $\quad i^{2}=-1$
$ 9+24i-16+6(-3-4i)+25=0\qquad$...Apply the distributive property: $\quad a(b-c)=ab-ac$
$ 9+24i-16+6(-3)+6(-4i)+25=0\qquad$...Simplify.
$-7+24i-18-24i+25=0\qquad$...add like terms.
$0=0$