$$(5x^2+4x)^2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 4 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (5x^2+4x)^2 &= 0&& \text{simplify left side} \\[1 em]25x^4+40x^3+16x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 25x^{4}+40x^{3}+16x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 25x^{4}+40x^{3}+16x^{2} = x^2 \left( 25x^{2}+40x+16 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 25x^{2}+40x+16 = 0$.
$ 25x^{2}+40x+16 = 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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