Answer
The solutions are $-3+\sqrt{5}$ and $-3-\sqrt{5}$
Work Step by Step
$ x^{2}+6x+4=0\qquad$ ... write left side in the form $x^{2}+bx$ (add $-4$ to each side).
$ x^{2}+6x+4-4=0-4\qquad$ ...simplify.
$ x^{2}+6x=-4\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=-4+9\qquad$ ... write left side as a binomial squared.
$(x+3)^{2}=-4+9\qquad$ ...simplify.
$(x+3)^{2}=5\qquad$ ...take square roots of each side.
$ x+3=\pm\sqrt{5}\qquad$ ...add $-3$ to each side
$ x+3-3=\pm\sqrt{5}-3\qquad$ ...simplify.
$x=-3\pm\sqrt{5}$
