Trig Help!!! DESPERATE!!?

Question

I need help on a problem. It is √3y+10 - 2 = y

If you cant tell the square root covers the 3y+10 and then you subtract two, and its equal to y......i really need help on this one.....

Answer 1

1) 3y+10>=0 so y>= -10/3
2)y+2>=0 so y>=-2 so this is the condition for existence
squaring
3y+10 = y^2+4y+4
so y^2+y-6=0 y=((-1+-sqrt(25))/2 y= 2 and y= -3(not a solutionj)
so y=2

Answer 2

√(3y+10) - 2 = y
=> √(3y+10) = y + 2, y ≥ -2 (domain)
Square both sides,
3y + 10 = y^2 + 4y + 4
=> y^2 + y - 6 = (y+3)(y-2) = 0
y = 2
y = -3 is not a solution since y ≥ -2.