Answer:
D. y =1 or y = -1
Step-by-step explanation:
We have the equation as following:
3. y^2 + 3y -6 =0
=> 3.y.y + 3.y - 6 = 0
=> 3.y.y + (6 - 3).y - 6 = 0
=> 3.y.y + 6y - 3y - 6= 0
=> 3.y.y + 2.3.y - (3y + 6) =0
=> 3y (y + 2) - 3(y + 2) = 0
=> (y + 2) ( 3y - 3) = 0
The above equation happens only when y + 2 = 0 or 3y -3 = 0
If y + 2 = 0
=> y = -2
If 3y -3 =0
=> 3y = 3
=> y = 3/3 = 1
So that the solutions for the equation are y = 1 or y = -2
