Answer
The solutions are $9+ i\sqrt{5}$ and $9-i\sqrt{5}$.
Work Step by Step
$ x^{2}-18x+86=0\qquad$ ... write left side in the form $x^{2}+bx$ (add $-86$ to each side).
$ x^{2}-18x+86-86=0+-86\qquad$ ...simplify.
$ x^{2}-18x=-86\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-18}{2})^{2}=9^{2}=81\qquad$ ...add $81$ to each side of the expression
$ x^{2}-18x+81=-86+81\qquad$ ...simplify.
$ x^{2}-18x+81=-5\qquad$ ... write left side as a binomial squared.
$(x-9)^{2}=-5\qquad$ ...take square roots of each side.
$ x-9=\pm\sqrt{-5}\qquad$ ...simplify.($\sqrt{-5}=i\sqrt{5}$)
$ x-9=\pm i\sqrt{5}\qquad$ ...add $9$ to each side.
$ x-9+9=\pm i\sqrt{5}+9\qquad$ ...simplify.
$x=9\pm i\sqrt{5}$
