Question
$$(6x^4+15x^3)(6x^3-15x^4) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 2 }{ 5 } & x_3 = -\dfrac{ 5 }{ 2 } \end{matrix} $$
Explanation
$$ \begin{aligned} (6x^4+15x^3)(6x^3-15x^4) &= 0&& \text{simplify left side} \\[1 em]36x^7-90x^8+90x^6-225x^7 &= 0&& \\[1 em]-90x^8-189x^7+90x^6 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -90x^{8}-189x^{7}+90x^{6} = 0 } $, first we need to factor our $ x^6 $.
$$ -90x^{8}-189x^{7}+90x^{6} = x^6 \left( -90x^{2}-189x+90 \right) $$$ x = 0 $ is a root of multiplicity $ 6 $.
The remaining roots can be found by solving equation $ -90x^{2}-189x+90 = 0$.
$ -90x^{2}-189x+90 = 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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