Answer
The solutions are $\frac{8}{3}$ and -9.
Work Step by Step
3$x^{2}$+19x-72=0
3$x^{2}$+27x-8x-72=0
3x(x+9)-8(x+9)=0
(3x-8)(x+9)=0
(3x-8)=0 or (x+9)=0
3x-8+8=0 +8 or x+9-9=0-9
3x=8 or x=-9
$\frac{3x}{3}$ =$\frac{8}{3}$ or x=-9
x= $\frac{8}{3}$ or x=-9
The solutions are $\frac{8}{3}$ and -9.
Check
Let x=$\frac{8}{3}$
3$x^{2}$+19x-72=0
3$(8/3)^{2}$+19$\frac{8}{3}$-72=0
3$\frac{64}{9}$+$\frac{152}{3}$-72=0
$\frac{64}{3}$+$\frac{152}{3}$-72=0
$\frac{216}{3}$ -72=0
72-72=0
0=0
Let x=-9
3$x^{2}$+19x+19(-9)-72=0
3$(-9)^{2}$ +19(-9)-72=0
3*81-171-72=0
243-171-72=0
0=0
