$$(2x^{11}-5x^7-10x^6)\cdot2x^3 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 1.54727 & x_3 = -1.06819+0.60246i \\[1 em] x_4 = -1.06819-0.60246i & x_5 = 0.29455+1.43592i & x_6 = 0.29455-1.43592i \end{matrix} $$
Explanation
$$ \begin{aligned} (2x^{11}-5x^7-10x^6)\cdot2x^3 &= 0&& \text{simplify left side} \\[1 em]4x^{14}-10x^{10}-20x^9 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 4x^{14}-10x^{10}-20x^{9} = 0 } $, first we need to factor our $ x^9 $.
$$ 4x^{14}-10x^{10}-20x^{9} = x^9 \left( 4x^{5}-10x-20 \right) $$$ x = 0 $ is a root of multiplicity $ 9 $.
The remaining roots can be found by solving equation $ 4x^{5}-10x-20 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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