$$-x^2(2x+8)(x+3) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -3 & x_3 = -4 \end{matrix} $$
Explanation
$$ \begin{aligned} -x^2(2x+8)(x+3) &= 0&& \text{simplify left side} \\[1 em]-(2x^3+8x^2)(x+3) &= 0&& \\[1 em]-(2x^4+6x^3+8x^3+24x^2) &= 0&& \\[1 em]-(2x^4+14x^3+24x^2) &= 0&& \\[1 em]-2x^4-14x^3-24x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -2x^{4}-14x^{3}-24x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ -2x^{4}-14x^{3}-24x^{2} = x^2 \left( -2x^{2}-14x-24 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ -2x^{2}-14x-24 = 0$.
$ -2x^{2}-14x-24 = 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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