\(2\left(x^2-3x+3,5\right)=2\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}\)
\(\left(x^2-2x+2\right)+\left(x^2-4x+5\right)=2\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}\)
\(\left(\sqrt{x^2-2x+2}-\sqrt{x^2-4x+5}\right)^2=0\)
x2-2x+2=x2-4x+5
=>2x = 3
=> \(x=\frac{3}{2}\)