Đáp án:
$x(x+1)+x^2=0\\↔x(x+1+x)=0\\↔x(2x+1)=0\\↔\left[ \begin{array}{l}x=0\\2x+1=0\end{array} \right.\\↔\left[ \begin{array}{l}x=0\\2x=-1\end{array} \right.\\↔\left[ \begin{array}{l}x=0\\x=-\dfrac{1}{2}\end{array} \right.\\Vậy\,\, S=\{0,-\dfrac{1}{2}\}$