Answer
$-3$ and $3$
Work Step by Step
Divide both sides of the equation by 4:
$4x^{2}-36=0\qquad.../\div 4$
$x^{2}-9=0$
... recognize a difference of squares (of x and 3)
$A^{2}-B^{2}=(A+B)(A-B)$
$(x+3)(x-3)=0$
By the zero product principle,
$x+3=0$ or $x-3=0$
$x=-3$ or $x=3$
The solutions are $-3$ and $3$.
