#### Answer

$\frac{-3}{2}$ and $\frac{3}{2}$

#### Work Step by Step

4$x^{2}$-9=0
$(2x)^{2}$-$3^{2}$=0
(2x+3)(2x-3)=0
2x+3=0 or 2x-3=0
2x=-3 or 2x=3
x=$\frac{-3}{2}$ or x= $\frac{3}{2}$
The solutions are
$\frac{-3}{2}$ and $\frac{3}{2}$
Check
Let x=$\frac{-3}{2}$
4$x^{2}$ -9=0
4$(-\frac{3}{2})^{2}$ -9=0
4$\frac{9}{4}$-9=0
9-9=0
0=0
Let x= $\frac{3}{2}$
4$x^{2}$ -9=0
4$(\frac{3}{2})^{2}$ -9=0
4$\frac{9}{4}$-9=0
9-9=0
0=0