Find the real solutions of sq. root(3x-5) minus sq.root(x+7) equals 2. Please explain. Thanks!?

Answer 1

sqrt(3x-5) - sqrt(x+7) = 2
sqrt(3x-5) = 2 - sqrt(x+7)
3x-5 = 4 -4sqrt(x+7) +x+7
2x -16 = -4sqrt(x+7)
4x^2 -64x +256 = 16x +112
4x^2-80x +144 = 0
x^2 - 20x +36 =0
(x-18)(x-2) = 0
x = 18 and x = 2
Checking these results shows that only x = 18 is correct