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