$x^4-4x^2-x+2=0$
$⇔(x^4+x^3)-(x^3+x^2)-(3x^2+3x)+(2x+2)=0$
$⇔(x+1)(x^3-x^2-3x+2)=0$
$⇔(x+1)[(x^3-2x^2)+(x^2-2x)-(x-2)]=0$
$⇔(x+1)(x-2)(x^2+x-1)=0$
$⇔\left[ \begin{array}{l}x=-1\\x=2\\x=\frac{\\-1±\sqrt{5}}{2}\end{array} \right.$
Vậy $S=\{-1;2;\frac{\\-1±\sqrt{5}}{2}\}$