Answer
$x=3+\sqrt{2}i$ or $x=3-\sqrt{2}i$
Work Step by Step
$ x^{2}+11=6x\qquad$...add $-6x$ to both sides.
$ x^{2}+11-6x=0\qquad$... solve with the Quadractic formula. $a=1,\ b=11,\ c=-6$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-6,\ a$ for $1$ and $c$ for $11$.
$ x=\displaystyle \frac{-(-6)\pm\sqrt{(-6)^{2}-4\cdot(11)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$x=\displaystyle \frac{6\pm\sqrt{36-44}}{2}$
$x=\displaystyle \frac{6\pm\sqrt{-8}}{2}$
$ x=\displaystyle \frac{6\pm\sqrt{-1\cdot 4\cdot 2}}{}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$)
$x=\displaystyle \frac{6\pm 2\sqrt{2}i}{2}$
$ x=3\pm\sqrt{2}i\qquad$... the symbol $\pm$ indicates two solutions.
$x=3+\sqrt{2}i$ or $x=3-\sqrt{2}i$