$$3x^5+2x^4-6x^3+8x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 0.99643+0.64862i & x_3 = 0.99643-0.64862i \\[1 em] x_4 = -1.32976+0.3438i & x_5 = -1.32976-0.3438i \\[1 em] \end{matrix} $$
Explanation
In order to solve $ \color{blue}{ 3x^{5}+2x^{4}-6x^{3}+8x = 0 } $, first we need to factor our $ x $.
$$ 3x^{5}+2x^{4}-6x^{3}+8x = x \left( 3x^{4}+2x^{3}-6x^{2}+8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{4}+2x^{3}-6x^{2}+8 = 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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