How to solve $(x^2-2x)^2+4(x^2-2x)+3=0$? How to solve $(x^2-2x)^2+4(x^2-2x)+3=0\,$? I tried
$$(x^2-2x)(x^2-2x+4)+3=0 \Longrightarrow (x^2-2x)(x^2-2x+4)=-3,$$ but couldn't continue. So that is why I came here to ask for some help, please. Thank you.
Lorie  Kirchner

$$\begin{align}(x^2-2x)^2+4(x^2-2x)+3 & =(x^2-2x)^2+4(x^2-2x)+4-1 \\\,\\
& =(x^2-2x+2)^2-1 \\\,\\ 
& =(x^2-2x+1)(x^2-2x+3)\\\,\\ 
& =(x-1)^2(x^2-2x+3) 
\end{align}$$
