Answer
The solutions are - $\frac{4}{3}$ and 5.
Work Step by Step
6$y^{2}$-22y-40=0
6$y^{2}$+8y-30y-40=0
2y(3y+4)-10(3y+4)=0
(2y-10)(3y+4)=0
2y-10 =0 or 3y+4=0
2y=10 0r y=-$\frac{4}{3}$
y=5 or y=-$\frac{4}{3}$
The solutions are - $\frac{4}{3}$ and 5.
Check
Let y=- $\frac{4}{3}$
6$y^{2}$-22y-40=0
6$(-4/3)^{2}$-22$\frac{-4}{3}$-40=0
6*16/9+$\frac{88}{3}$-40=0
2*16/3+$\frac{88}{3}$-40=0
32/3-$+frac{88}{3}$-40=0
$\frac{120}{3}$-40=0
40-40=0
0=0
Let y=5
6$y^{2}$-22y-40=0
6$(5)^{2}$-22*5-40=0
6*25-110-40=0
150-110-40=0
0=0
