用费拉里法解方程的思路，是将一个一元四次方程，拆分成两个一元二次方程
也就是

$$
\large
\begin{align*}
& x4+bx3+cx^2+dx+e\newline
= & (x2+s_{1}x+t_{1})(x2+s_{2}x+t_{2})\newline
= & 0
\end{align*}
$$

解这2个方程$ x2+s_{1}x+t_{1}=0,x2+s_{2}x+t_{2}=0 $，即可得到一元四次方程的4个根
令

$$
\large
\begin{array}{l}
p= bd-4e-\dfrac{c^2}{3}\newline
q=e(4c-b2)-d2+\dfrac{c}{3}(bd-4e)-\dfrac{2}{27}c^3\newline
\Delta=\dfrac{q^2}{4} +\dfrac{p^3}{27}
\end{array}
$$

一、$ \Delta \geqslant 0 $时
$$
\large
\begin{align*}
\begin{array}{l}
y=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\Delta}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\Delta}}+\dfrac{c}{3}\newline
s_{1}=\dfrac{1}{2}\left( b+\sqrt{4y+b^2-4c} \right)\newline
t_{1}=\dfrac{y}{2}+\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} }\newline
s_{2}=\dfrac{1}{2}\left( b-\sqrt{4y+b^2-4c} \right)\newline
t_{2}=\dfrac{y}{2}-\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} }
\end{array}
\end{align*}
$$
至此，已将方程分解成了2个一元二次方程
$$
\large
\begin{align*}
\begin{array}{l}
x^2+s_{1}x+t_{1}=0\newline
x^2+s_{2}x+t_{2}=0
\end{array}
\end{align*}
$$
解这两个一元二次方程得到的4个根，即为一元四次方程的根
4个根分别为
$$
\large
\begin{align*}
\begin{array}{l}
x_{1,2}=\dfrac{-s_{1} \pm \sqrt{s_{1}^{2}-4t_{1}}}{2}\newline
x_{3,4}=\dfrac{-s_{2} \pm \sqrt{s_{2}^{2}-4t_{2}}}{2}
\end{array}
\end{align*}
$$
二、$ \Delta<0 $ 时
$ \Delta=\dfrac{q^2}{4} +\dfrac{p^3}{27}<0 $，且$ p,q $都是实数，可得出$ p<0 $
$$
\large
\begin{align*}
\begin{array}{l}
T=\dfrac{q}{2p}\sqrt{-\dfrac{27}{p}}\newline
\theta=\arccos T\newline
y=2\sqrt{-\dfrac{p}{3}}\cos\left(\dfrac{\theta}{3}\right)+\dfrac{c}{3}\newline
s_{1}=\dfrac{1}{2}\left( b+\sqrt{4y+b^2-4c} \right)\newline
t_{1}=\dfrac{y}{2}+\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} }\newline
s_{2}=\dfrac{1}{2}\left( b-\sqrt{4y+b^2-4c} \right)\newline
t_{2}=\dfrac{y}{2}-\dfrac{0.5by-d}{\sqrt{4y+b^2-4c} }
\end{array}
\end{align*}
$$
方程的4个根分别为
$$
\large
\begin{align*}
\begin{array}{l}
x_{1,2}=\dfrac{-s_{1} \pm \sqrt{s_{1}^{2}-4t_{1}}}{2}\newline
x_{3,4}=\dfrac{-s_{2} \pm \sqrt{s_{2}^{2}-4t_{2}}}{2}
\end{array}
\end{align*}
$$