Quadratic equation
Solve=
square root(1-y) + square root(12+y) -5 =0
√(1-y) + √(12+y) - 5 = 0
As it is written, it is not a quadratic equation.
√(1-y) = 5 - √(12+y)
square both sides to get
1-y = 25 - 10√(12+y) + 12+y
10√(12+y) = 36 + 2y
square again:
100(12+y) = 1296 + 144y + 4y^2
Now we finally have a quadratic equation we can work with!
4y^2 + 44y + 96 = 0
y^2 + 11y + 24 = 0
(y+3)(y+8) = 0
y = -3 or -8
But, with all the squaring, we have almost certainly introduced extraneous solutions. So, check in the original equation:
√(1+3) + √(12-3) - 5 = 2+3-5 = 0 OK
√(1+8) + √(12-8) - 5 = 3+2-5 = 0 OK