Answer
The solutions are $-2+\sqrt{14}$ and $-2-\sqrt{14}$.
Work Step by Step
$ x^{2}+4x=10\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{4}{2})^{2}=2^{2}=4\qquad$ ...add $4$ to each side of the expression
$ x^{2}+4x+4=10+4\qquad$ ... write left side as a binomial squared.
$(x+2)^{2}=10+4\qquad$ ...simplify.
$(x+2)^{2}=14\qquad$ ...take square roots of each side.
$ x+2=\pm\sqrt{14}\qquad$ ...add $-2$ to each side
$ x+2-2=\pm\sqrt{14}-2\qquad$ ...simplify.
$x=-2\pm\sqrt{14}$
