$$-2x^2+x^5-4x+3x^7 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -0.89673 & x_3 = 1.08131 \\[1 em] x_4 = 0.46025+1.03356i & x_5 = 0.46025-1.03356i & x_6 = -0.55254+0.87687i \\[1 em] x_7 = -0.55254-0.87687i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} -2x^2+x^5-4x+3x^7 &= 0&& \text{simplify left side} \\[1 em]3x^7+x^5-2x^2-4x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{7}+x^{5}-2x^{2}-4x = 0 } $, first we need to factor our $ x $.
$$ 3x^{7}+x^{5}-2x^{2}-4x = x \left( 3x^{6}+x^{4}-2x-4 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{6}+x^{4}-2x-4 = 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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