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(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