Answer
$x=-3+\sqrt{19}$ or $x=-3-\sqrt{19}$
Work Step by Step
$x^{2}+6x=10$
$\qquad$...square half the coefficient of $x$ and add it to both sides to complete the square
$(\displaystyle \frac{6}{2})=3,\ 3^{2}=9$
$ x^{2}+6x+9=19\qquad$...write $x^{2}+6x+9$ as a binomial squared.
$(x+3)^{2}=19\qquad$...use the general principle of square roots:
For any real number $k$ and any algebraic expression $X$ :
$\text{If }X^{2}=k,\text{ then }X=\sqrt{k}\text{ or }X=-\sqrt{k}$
$ x+3=\pm\sqrt{19}\qquad$...add $-3$ to both sides.
$ x=-3\pm\sqrt{19}\qquad$... the symbol $\pm$ indicates two solutions.
$x=-3+\sqrt{19}$ or $x=-3-\sqrt{19}$