Solve the following quartic equation:

$$
(x-1)(x-2)(x-3)(x-4)=120 .
$$
If I multiply the first factor by the last and the second by the third, the equation takes the following form.

$$
\left(x^{2}-5 x+4\right)\left(x^{2}-5 x+6\right)=120
$$

Or

$$
\begin{gathered}
{\left[\left(x^{2}-5 x+5\right)-1\right]\left[\left(x^{2}-5 x+5\right)+1\right]=120} \\
\left(x^{2}-5 x+5\right)^{2}-1=120
\end{gathered}
$$

From this

$$
x^{2}-5 x+5= \pm \sqrt{121}
$$

which yields two specific quadratic equations:

These are

$$
\begin{gathered}
x^{2}-5 x-6=0 \\
x^{2}-5 x+16=0
\end{gathered}
$$

The roots of the first are:

$$
x'=-1, \quad x''=6
$$

and those of the second are:

$$
x=\frac{5 \pm \sqrt{-39}}{2}
$$