Answer
The solutions are $x=5+2\sqrt{7}$ and $x=5-2\sqrt{7}$.
Work Step by Step
$ 4x^{2}-40x-12=0\qquad$ ...divide each term with $4$.
$ x^{2}-10x-3=0\qquad$ ...add $3$ to each side.
$ x^{2}-10x-3+3=0+3\qquad$ ...simplify.
$ x^{2}-10x=3\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-10}{2})^{2}=(-5)^{2}=25\qquad$ ...add $25$ to each side of the expression
$ x^{2}-10x+25=3+25\qquad$ ...simplify.
$ x^{2}-10x+25=28\qquad$ ... write left side as a binomial squared.
$(x-5)^{2}=28\qquad$ ...take square roots of each side.
$ x-5=\pm\sqrt{28}\qquad$ ...rewrite $\sqrt{28}$ as $\sqrt{4\cdot 7}$
$ x-5=\pm\sqrt{4\cdot 7}\qquad$ ...evaluate $\sqrt{4}$.
$ x-5=\pm 2\sqrt{7}\qquad$ ...add $5$ to each side.
$x=5\pm 2\sqrt{7}$
