$$9x^3(7x^2+5x-2)+11x^2(6x^2-5x+5) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -2.39719 & x_3 = 0.31764+0.51311i \\[1 em] x_4 = 0.31764-0.51311i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 9x^3(7x^2+5x-2)+11x^2(6x^2-5x+5) &= 0&& \text{simplify left side} \\[1 em]63x^5+45x^4-18x^3+66x^4-55x^3+55x^2 &= 0&& \\[1 em]63x^5+111x^4-73x^3+55x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 63x^{5}+111x^{4}-73x^{3}+55x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 63x^{5}+111x^{4}-73x^{3}+55x^{2} = x^2 \left( 63x^{3}+111x^{2}-73x+55 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 63x^{3}+111x^{2}-73x+55 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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