\(x^4=2x^2+8x+3\)
\(\Leftrightarrow\left(x^4+2x^2+1\right)+\left(-4x^2-8x-4\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2-2x-1\right)=0\)
Dễ thấy \(x^2+2x+3>0\)
\(\Rightarrow x^2-2x-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{cases}}\)