The given equation is $$x(x^2+3|x|+5|x-1|+6|x-2|)=0$$, which can be solved by analyzing it in parts. It can be broken down into: $$x=0$$ and $$x^2+3|x|+5|x-1|+6|x-2|=0$$.

For $$x=0$$, it's clear that it is a solution to the equation since it makes the entire expression equal to zero.

Case (I)

$$ x<0 $$

$$ \begin{aligned} & x^2-3 x-5(x-1)-6(x-2)=0 \\\\ & x^2-14 x+17=0 \end{aligned} $$

$\because$ All roots are positive $\Rightarrow$ no solution

Case (II)

$$ \begin{aligned} & 0 < x < 1 \\\\ & x^2+3 x-5(x-1)-6(x-2)=0 \\\\ & x^2-8 x+17=0 \\\\ & \because D < 0 \Rightarrow \text { no solution } \end{aligned} $$

Case (III)

$$ 1 < x < 2 $$

$$ x^2+3 x+5(x-1)-6(x-2)=0 $$

$$ x^2+2 x+7=0 $$

$\Rightarrow$ no solution

Case (IV)

$$ \begin{aligned} & x > 2 \\\\ & x^2+3 x+5(x-1)+6(x-2)=0 \\\\ & x^2+14 x-19=0 \end{aligned} $$

All roots less than 2

$\Rightarrow$ no solution

Here $x=0$ is only solution.