\(x^4+x^2+6x-8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
\(\Leftrightarrow x=1\left(h\right)x=-2\left(h\right)x^2-x+4=0\)
Mà \(x^2-x+4=\left(x-\frac{1}{2}\right)^2+\frac{15}{4}>0\)
\(\Rightarrow x=1\left(h\right)x=-2\)