Solve I x^2-4 I = 2-x algebraically
|x^2-4| = 2-x
since absolute values are never negative, we know that x <= 2
So now we have to solve
x^2-4 = 2-x
-(x^2-4) = 2-x
x^2+x-2 = 0 .. x = -2 or 1
x^2+x-6 = 0 .. x = -3 or 2
all of these fit the original equation
Or, solve by squaring.
|x^2-4| = 2-x
x^2-4 = ±(2-x)
(x^2-4)^2 = (2-x)^2
x^4 - 8x^2 + 16 = x^2 - 4x + 4
x^4 - 9x^2 + 4x + 12 = 0
(x+3)(x+1)(x-2)^2 = 0