$$4x^5+5x^3-3x^2+x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 0.29837+0.29261i & x_3 = 0.29837-0.29261i \\[1 em] x_4 = -0.29837+1.15863i & x_5 = -0.29837-1.15863i \\[1 em] \end{matrix} $$
Explanation
In order to solve $ \color{blue}{ 4x^{5}+5x^{3}-3x^{2}+x = 0 } $, first we need to factor our $ x $.
$$ 4x^{5}+5x^{3}-3x^{2}+x = x \left( 4x^{4}+5x^{2}-3x+1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 4x^{4}+5x^{2}-3x+1 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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