$$3x(x-4)(x+5) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 4 & x_3 = -5 \end{matrix} $$
Explanation
$$ \begin{aligned} 3x(x-4)(x+5) &= 0&& \text{simplify left side} \\[1 em](3x^2-12x)(x+5) &= 0&& \\[1 em]3x^3+15x^2-12x^2-60x &= 0&& \\[1 em]3x^3+3x^2-60x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{3}+3x^{2}-60x = 0 } $, first we need to factor our $ x $.
$$ 3x^{3}+3x^{2}-60x = x \left( 3x^{2}+3x-60 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{2}+3x-60 = 0$.
$ 3x^{2}+3x-60 = 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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