Ta có: (x+5)(3x-12)<0
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+5>0\\3x-12< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+5< 0\\3x-12>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-5\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x< -5\\x>4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-5< x< 4\)
Vậy: S={x|-5<x<4}