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