Question 782302
8y + 3y^2 = 35
3y^2 + 8y - 35 = 0
use quadratic equation to solve:

(-8 +-sqrt8^2 - 4(3)(-35))/2(3)
(-8 +-sqrt64 +420) /6
(-8 +-sqrt484) /6
(-8 +- 22) /6
(-4 +- 11) /3

x = (-4 + 11) /3 = 2.33

x = (-4 - 11) /3 = -5