Correct answer: (-1/2, 1), x != -2
Numerator = (x+2)² >= 0, zero only at x = -2. Denominator = 2x² - x - 1 = (2x+1)(x-1), roots at x = -1/2 and x = 1. The fraction is < 0 only when numerator > 0 and denominator < 0. Numerator > 0 for x != -2. Denominator < 0 between its roots: -1/2 < x < 1. Exclude x = -2 (it lies outside this interval anyway, but in general the numerator zero must be excluded). So solution = (-1/2, 1) with x != -2; since -2 is not in (-1/2,1) the set is effectively (-1/2, 1). The form listed as the answer keeps the x != -2 condition explicit.