Answer
The solutions are $\frac{-3}{2}$ and $\frac{5}{4}$.
Work Step by Step
(2x+3)(4x-5)=0
2x+3=0 or 4x-5=0
2x+3-3=0-3 or 4x-5+5=0+5
2x=-3 or 4x=5
$\frac{2x}{2}$=$\frac{-3}{2}$ or $\frac{4x}{4}$=$\frac{5}{4}$
x=-$\frac{3}{2}$ or x=$\frac{5}{4}$
The solutions are $\frac{-3}{2}$ and $\frac{5}{4}$.
check
let x=-$\frac{3}{2}$
(2x+3)(4x-5)=0
[2(-$\frac{3}{2}$ ) +3][4 (-$\frac{3}{2}$ )-5]=
(-3+3) [4 (-$\frac{3}{2}$ )-5]=0
0[4 (-$\frac{3}{2}$ )-5]=0
0=0
let x=$\frac{5}{4}$
(2x+3)(4x-5)=0
(2$\frac{5}{4}$ +3)(4 $\frac{5}{4}$ -5)=0
(2$\frac{5}{4}$ +3) (5-5)=0
(2$\frac{5}{4}$ +3) 0=0
0=0
