122. Without expanding the brackets. Solve the following equation without expanding the brackets:

$$
(12 x-1)(6 x-1)(4 x-1)(3 x-1)=5 \text{. }
$$
122. Substituting $x=\frac{y}{12}$ into the original equation, we get

$$
(y-1)(y-2)(y-3)(y-4)=120=2 \cdot 3 \cdot 4 \cdot 5
$$

Any integer root of this equation will turn its left side into the product of four consecutive integers. $y=-1$ and $y=6$ are two such roots.

Now let's use the known relationships between the roots of the equation and its coefficients

$$
(-1)(6) r_{1} r_{2}=-120+(-1)(-2)(-3)(-4)
$$

from which

$$
\begin{aligned}
r_{1} r_{2} & =16 \\
-\left(-1+6+r_{1}+r_{2}\right) & =-1-2-3-4
\end{aligned}
$$

from which

$$
r_{1}+r_{2}=5
$$

Thus, $r_{1}$ and $r_{2}$ coincide with the roots of the equation $y^{2}-5 y+16=0$, that is, they are equal to $\frac{(5 \pm i \sqrt{39})}{2}$.

Therefore, the four roots of the original equation are respectively $-\frac{1}{12}, \frac{1}{2}$ and $\frac{5 \pm i \sqrt{39}}{24}$.