$x^4+2x^2-7x-10=0$
$⇔(x^4+x^3)-(x^3+x^2)+(3x^2+3x)-(10x+10)=0$
$⇔(x+1)(x^3-x^2+3x-10)=0$
$⇔(x+1)[(x^3-2x^2)+(x^2-2x)+(5x-10)]=0$
$⇔(x+1)(x-2)(x^2+x+5)=0$
$⇔(x+1)(x-2)[(x+\frac{1}{2})^2+\frac{19}{4}]=0$
Vì $(x+\frac{1}{2})^2≥0∀x$ nên $(x+\frac{1}{2})^2+\frac{19}{4}>0∀x$
$⇒(x+1)(x-2)=0$
$⇔\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.⇔\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.$
Vậy $S=\{-1;2\}$.
