$$(3x-1)(2x^2+3x) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 3 } & x_3 = -\dfrac{ 3 }{ 2 } \end{matrix} $$
Explanation
$$ \begin{aligned} (3x-1)(2x^2+3x) &= 0&& \text{simplify left side} \\[1 em]6x^3+9x^2-2x^2-3x &= 0&& \\[1 em]6x^3+7x^2-3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 6x^{3}+7x^{2}-3x = 0 } $, first we need to factor our $ x $.
$$ 6x^{3}+7x^{2}-3x = x \left( 6x^{2}+7x-3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 6x^{2}+7x-3 = 0$.
$ 6x^{2}+7x-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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