Quartic equations are in the form of \[ax^{4}+bx^{3}+cx^{2}+dx+e=0\] or
\begin{align*}
\tag {1}
x^{4}+\frac{b}{a}x^{3}+\frac{c}{a}x^{2}+\frac{d}{a}x+\frac{e}{a}=0
\end{align*}
Let the roots of the equation above be \[\alpha\], \[\beta\], \[\gamma\] and \[\delta\].
\begin{align*}
(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)&=0 \\
\tag {2}
x^{4}-(\alpha+\beta+\gamma+\delta)x^{3}&+(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)x^{2}\\
&-(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta)x+\alpha\beta\gamma\delta=0
\end{align*}
By comparing (1) and (2) we get that:
\begin{align*}
\alpha+\beta+\gamma+\delta&=-\frac{b}{a}=\Sigma\alpha=S_{1} \\
\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta&=\frac{c}{a}=\Sigma\alpha\beta \\
\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta&=-\frac{d}{a}=\Sigma\alpha\beta\gamma \\
\alpha\beta\gamma\delta&=\frac{e}{a}=\Sigma\alpha\beta\gamma\delta \\
\end{align*}
As for \[S_{2}\], the formula is identical to cubics and quadratics, where \[S_{2}=(\Sigma\alpha)^2-2(\Sigma\alpha\beta)\]
NOTE: Due to the length of the expansion of these equation, it's much more sensible to find \[S_{3}\] by taking \[aS_{3}=-(bS_{2}+cS_{1}+dS_{0}+eS_{-1})\].