$x³ +3x² +x² -9 = 0$
$⇔ x².(x +3) +(x -3).(x +3) = 0$
$⇔ (x +3).(x² +x -3) = 0$
$⇔ (x +3).[(x + \frac{1}{2})² - \frac{13}{4}] = 0$
$⇔ (x +3).(x + \frac{1}{2} - \frac{√13}{2}).(x + \frac{1}{2} + \frac{√13}{2}) = 0$
$⇔ \left[ \begin{array}{l}x +3=0\\x + \frac{1}{2} - \frac{√13}{2}=0\\x + \frac{1}{2} + \frac{√13}{2}) = 0\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=-3\\x=\frac{-1 +√13}{2}\\x = \frac{-1 -√13}{2}\end{array} \right.$
$Vậy$ $S =$ {$-3; \frac{-1 +√13}{2}; \frac{-1 -√13}{2}$}
