Answer
The solutions are $\frac{2}{3}$ and $\frac{-1}{5}$.
Work Step by Step
(3x-2)(5x+1)=0
3x-2=0 or 5x+1=0
3x-2+2=0+2 or 5x+1-1=0-1
3x=2 or 5x=-1
$\frac{3x}{3}$=$\frac{2}{3}$ or $\frac{5x}{5}$=$\frac{-1}{5}$
x=$\frac{2}{3}$ or x=$\frac{-1}{5}$
The solutions are $\frac{2}{3}$ and $\frac{-1}{5}$.
check
let x=$\frac{-1}{5}$
(3x-2)(5x+1)=0
[3(-$\frac{1}{5}$ ) -2][5 (-$\frac{1}{5}$ )+1]=
[3(-$\frac{1}{5}$ ) -2] [-1+1]]=0
[3(-$\frac{1}{5}$ ) -2]0=0
0=0
let x=$\frac{2}{3}$
(3x-2)(5x+1)=0
(3$\frac{2}{3}$ -2)(5$\frac{2}{3}$ +1)=0
(3$\frac{2}{3}$ -2)(5$\frac{2}{3}$ +1 =0
(2-2)(5$\frac{2}{3}$ +1)=0
0=0
