$$2g^2(3g^2-2) = 0$$
Answer
$$ \begin{matrix}g_1 = 0 & g_2 = - \dfrac{\sqrt{ 6 }}{ 3 } & g_3 = \dfrac{\sqrt{ 6 }}{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} 2g^2(3g^2-2) &= 0&& \text{simplify left side} \\[1 em]6g^4-4g^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 6x^{4}-4x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 6x^{4}-4x^{2} = x^2 \left( 6x^{2}-4 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 6x^{2}-4 = 0$.
$ 6x^{2}-4 = 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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