$36x^2-12x-15=0$ solve for $x$

Question

$\begin{array}{1 1} (a)\;\frac{1}{2},\frac{1}{3} \\ (b)\;\frac{-1}{2},\frac{-1}{3} \\ (c)\;\frac{-1}{2},\frac{5}{6} \\(d)\;\frac{1}{2},\frac{5}{6}\end{array}$