Ta có: \(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x^2+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\x^2=-4\\x=1\end{matrix}\right.\) (loại) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{5}\\x=1\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{-3}{5};1\right\}\)