ОДЗ: x+2>0 x-6>0
x> -2 x>6
x∈(6; +∞)
(√(x+2) - √(x-6))²=2²
x+2-2√[(x+2)(x-6)]+x-6=4
2x-4-2√(x²+2x-6x-12)=4
-2√(x²-4x-12)=4+4-2x
-2√(x²-4x-12)=8-2x
2√(x²-4x-12)=2x-8
√(x²-4x-12)=x-4
x²-4x-12=(x-4)²
x²-4x-12=x²-8x+16
x²-x²-4x+8x=16+12
4x=28
x=7
Ответ: 7.