+128475 f(f(x)) + 2x - 1 = (x^2 -2x)^2 - 2 (x^2 - 2x) + 2x - 1 = [ x^4 - 4x^3 + 4x^2 ] - 2x^2 + 4x + 2x- 1 =
x^4 - 4x^3 + 2x^2 + 6x - 1
So
x^2 - 2x = x^4 - 4x^3 + 2x^2 + 6x - 1 rearrange as
x^4 - 4x^3 + x^2 + 8x - 1 = 0
See the graph here : https://www.desmos.com/calculator/os2uyqpfze
The real numbers that make this true are x ≈ -1.209 and x ≈ .124
