(x^2 + 2x - 12) + √(x^2 + 2x + 8) = 0
(x^2 + 2x + 8) + √(x^2 + 2x + 8) - 20 = 0
Замена: √(x^2 + 2x + 8) = t ≥ 0
t^2 + t - 20 = 0
D=81
t1 = 4, t2 < 0
√(x^2 + 2x + 8) = 4
x^2 + 2x + 8 ≥ 0 при любых x
x^2 + 2x + 8 = 16
x^2 + 2x - 8 = 0
D=4+4*8 = 36
x1 = (-2 - 6)/2 = -8/2 = -4
x2 = (-2 + 6)/2 = 4/2 = 2
Ответ: x = 2, x = -4