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

Due to Viëta we have $x^2-2x=-1$ or $x^2-2x=-3$.  Clearly $x=1$ is the only (real) solution.
