$$5x(x^2-4x-2) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 2-\sqrt{ 6 } & x_3 = 2+\sqrt{ 6 } \end{matrix} $$
Explanation
$$ \begin{aligned} 5x(x^2-4x-2) &= 0&& \text{simplify left side} \\[1 em]5x^3-20x^2-10x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 5x^{3}-20x^{2}-10x = 0 } $, first we need to factor our $ x $.
$$ 5x^{3}-20x^{2}-10x = x \left( 5x^{2}-20x-10 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 5x^{2}-20x-10 = 0$.
$ 5x^{2}-20x-10 = 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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