$$(3x-2)x(2x-4) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 2 & x_3 = \dfrac{ 2 }{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} (3x-2)x(2x-4) &= 0&& \text{simplify left side} \\[1 em](3x^2-2x)(2x-4) &= 0&& \\[1 em]6x^3-12x^2-4x^2+8x &= 0&& \\[1 em]6x^3-16x^2+8x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 6x^{3}-16x^{2}+8x = 0 } $, first we need to factor our $ x $.
$$ 6x^{3}-16x^{2}+8x = x \left( 6x^{2}-16x+8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 6x^{2}-16x+8 = 0$.
$ 6x^{2}-16x+8 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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