$$-3x^2(x+5)(3x-8) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -5 & x_3 = \dfrac{ 8 }{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} -3x^2(x+5)(3x-8) &= 0&& \text{simplify left side} \\[1 em]-(3x^3+15x^2)(3x-8) &= 0&& \\[1 em]-(9x^4-24x^3+45x^3-120x^2) &= 0&& \\[1 em]-(9x^4+21x^3-120x^2) &= 0&& \\[1 em]-9x^4-21x^3+120x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -9x^{4}-21x^{3}+120x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ -9x^{4}-21x^{3}+120x^{2} = x^2 \left( -9x^{2}-21x+120 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ -9x^{2}-21x+120 = 0$.
$ -9x^{2}-21x+120 = 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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