(x-5)² + (x-5) - 2 = 0

Let (x-5) = u. Substituting u for (x-5) gives:

u² + u - 2 = 0

Factor that as

(u + 2)(u - 1) = 0

u + 2 = 0;  u - 1 = 0
    u = -2;     u = 1

But we don't want u, we want x, so we
substitute (x-5) for u

(x-5) = -2;  (x-5) = 1
  x-5 = -2;    x-5 = 1
    x = 3;      x = 6

Edwin