Answer
The solutions are $-3+2\sqrt{3}$ and $-3-2\sqrt{3}$.
Work Step by Step
$ x^{2}+6x-3=0\qquad$ ... write left side in the form $x^{2}+bx$ (add $3$ to each side).
$ x^{2}+6x-3+3=0+3\qquad$ ...simplify.
$ x^{2}+6x=3\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{6}{2})^{2}=3^{2}=9\qquad$ ...add $9$ to each side of the expression
$ x^{2}+6x+9=3+9\qquad$ ...simplify.
$ x^{2}+6x+9=12\qquad$ ... write left side as a binomial squared.
$(x+3)^{2}=12\qquad$ ...take square roots of each side.
$ x+3=\pm\sqrt{12}\qquad$ ...rewrite $\sqrt{12}$ as $\sqrt{4\cdot 3.}$
$ x+3=\pm\sqrt{4\cdot 3.}\qquad$ ...evaluate $\sqrt{4}$.
$ x+3=\pm 2\sqrt{3}\qquad$ ...add $-3$ to each side.
$ x+3-3=\pm 2\sqrt{3}-3\qquad$ ...simplify.
$x=-3\pm 2\sqrt{3}$
