#### 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.
$0=0$