if so, then let y = √(x^2+x)
your equation becomes
y^2 + y - 2 = 0
(y+2)(y-1) = 0
y = -2 or y = 1
sub back into the substitution and you have two quadratic equations to solve.
√(x^2 + x) = -2 which of course cannot have a real number solution and
√(x^2+x) = 1
square both sides and proceed from there.