Answer
The solutions are $\frac{7}{2}$ and $\frac{-2}{7}$
Work Step by Step
(2x-7)(7x+2)=0
2x-7=0 or 7x+2=0
2x-7+7=0+7 or 7x+2-2=0-2
2x=7 or 7x=-2
$\frac{2x}{2}$=$\frac{7}{2}$ or $\frac{7x}{7}$=$\frac{-2}{7}$
x=$\frac{7}{2}$ or x=$\frac{-2}{7}$
The solutions are $\frac{7}{2}$ and $\frac{-2}{7}$.
check
Let x=$\frac{7}{2}$
(2$\frac{7}{2}$ -7)(7$\frac{7}{2}$ +2)=
(7-7))(7$\frac{7}{2}$ +2)=0
(0)(7$\frac{7}{2}$ +2)=0
0=0
Let x=$\frac{-2}{7}$
(2x-7)(7x+2)=0
[2($\frac{-2}{7}$)-7][7($\frac{-2}{7}$)+2]
[2($\frac{-2}{7}$)-7](-2+2)=0
[2($\frac{-2}{7}$)-7]0=0
0=0
