$$0 = 2x^3+3x$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{\sqrt{ 6 }}{ 2 } i & x_3 = - \dfrac{\sqrt{ 6 }}{ 2 } i \end{matrix} $$
Explanation
$$ \begin{aligned} 0 &= 2x^3+3x&& \text{move all terms to the left hand side } \\[1 em]0-2x^3-3x &= 0&& \text{simplify left side} \\[1 em]-2x^3-3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -2x^{3}-3x = 0 } $, first we need to factor our $ x $.
$$ -2x^{3}-3x = x \left( -2x^{2}-3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -2x^{2}-3 = 0$.
$ -2x^{2}-3 = 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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