ОДЗ
{x-2≥0⇒x≥2
{x+1≥0⇒x≥-1
{x+6≥0⇒x≥-6
x∈[2;∞)
возведем в квадрат
x-2+x+1+2√(x²-x-2)>x+6
2√(x²-x-2)>7-x
возведем в квадрат
4(x²-x-2)>(7-x)²
4x²-4x-8-49+14x-x²>0
3x²+10x-57>0
D=100+684=784
x1=(-10-28)/6=-19/3
x2=(-10+28)/6=3
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+ _ +
--------(-19/3)--------(2)------(3)-----------------
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Ответ х∈(3;∞)