Real solution of the equation $(x^2-2x+2)^2-2(x^2-2x+2)+2 = x$ 
Calculate all real solutions $x\in\mathbb{R}$ of the equation
  $$
\tag1(x^2-2x+2)^2-2(x^2-2x+2)+2 = x
$$

My Attempt:
I used the concept of a composite function. Let $f(x) = x^2-2x+2$. Then equation $(1)$ converts into $f(f(x)) = x$. Both $f(x) = x$ and $f(x) = -x$ satisfy the given composite function.
Case 1: If $f(x) = x$, then
$$
x^2-2x+2=x\\
x^2-3x+2=0\\
x\in\{1,2\}
$$
Note that $1,2\in\mathbb{R}$.
Case 2: If $f(x)=-x$, then
$$
x^2-2x+2=-x\\
x^2-x+2=0\\
x=\frac{1\pm \sqrt{1-8}}{2}\notin \mathbb{R}
$$
So only $x\in\{1,2\}$ are the real solutions of above equation.
Is my process correct? Is there is any other method by which we can solve the above question?

Setting $T=x-1$ you have $(T^2+1)^2-2(T^2+1)+1-T=0$ or $T^4-T=0$ that is
$T(T-1)(T^2+T+1)=0$ so for real solutions you have $T=0$ or $T=1$, that is $x=1$ or $x=2$.
