$$(6x^2-5x)(6x^2+5x) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 5 }{ 6 } & x_3 = -\dfrac{ 5 }{ 6 } \end{matrix} $$
Explanation
$$ \begin{aligned} (6x^2-5x)(6x^2+5x) &= 0&& \text{simplify left side} \\[1 em]36x^4+30x^3-30x^3-25x^2 &= 0&& \\[1 em]36x^4+30x^3-30x^3-25x^2 &= 0&& \\[1 em]36x^4-25x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 36x^{4}-25x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 36x^{4}-25x^{2} = x^2 \left( 36x^{2}-25 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 36x^{2}-25 = 0$.
$ 36x^{2}-25 = 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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