$$3x^5-87x^3-300x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 5.66691 & x_3 = -5.66691 \\[1 em] x_4 = 1.76463i & x_5 = -1.76463i \\[1 em] \end{matrix} $$
Explanation
In order to solve $ \color{blue}{ 3x^{5}-87x^{3}-300x = 0 } $, first we need to factor our $ x $.
$$ 3x^{5}-87x^{3}-300x = x \left( 3x^{4}-87x^{2}-300 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{4}-87x^{2}-300 = 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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