Question
$$-3yy(6y+yy-5) = 0$$
Answer
$$ \begin{matrix}y_1 = 0 & y_2 = -3-\sqrt{ 14 } & y_3 = -3+\sqrt{ 14 } \end{matrix} $$
Explanation
$$ \begin{aligned} -3yy(6y+yy-5) &= 0&& \text{simplify left side} \\[1 em]-3y^2(6y+y^2-5) &= 0&& \\[1 em]-(18y^3+3y^4-15y^2) &= 0&& \\[1 em]-18y^3-3y^4+15y^2 &= 0&& \\[1 em]-3y^4-18y^3+15y^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -3x^{4}-18x^{3}+15x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ -3x^{4}-18x^{3}+15x^{2} = x^2 \left( -3x^{2}-18x+15 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ -3x^{2}-18x+15 = 0$.
$ -3x^{2}-18x+15 = 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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