$$3x(2x^2+1)(x^2-3) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 1.73205 & x_3 = -1.73205 \\[1 em] x_4 = 0.70711i & x_5 = -0.70711i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 3x(2x^2+1)(x^2-3) &= 0&& \text{simplify left side} \\[1 em](6x^3+3x)(x^2-3) &= 0&& \\[1 em]6x^5-18x^3+3x^3-9x &= 0&& \\[1 em]6x^5-15x^3-9x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 6x^{5}-15x^{3}-9x = 0 } $, first we need to factor our $ x $.
$$ 6x^{5}-15x^{3}-9x = x \left( 6x^{4}-15x^{2}-9 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 6x^{4}-15x^{2}-9 = 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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