Solve the equation

$$
2\left(x^{2}-1\right)(x-2)(x-3)+3(x-1)\left(x^{2}-4\right)=0
$$
Let's transform the left side of our equation into a product by factoring out the common factor $(x-1)(x-2)$ from both terms:

$$
(x-1)(x-2)[2(x+1)(x-3)+3(x+2)]=0
$$

Expanding the brackets inside the square brackets:

$$
2 x^{2}-4 x-6+3 x+6=2 x^{2}-x
$$

Thus, our equation becomes:

$$
(x-1)(x-2) x(2 x-1)=0
$$

From here, the roots are:

$$
x_{1}=0, \quad x_{2}=1, \quad x_{3}=2, \quad x_{4}=\frac{1}{2}
$$

Varga Ferenc (Sárospatak, Rákóczi g. I. o. t.)