$$(2m^4+5m^2)(m^3-7m) = 0$$
Answer
$$ \begin{matrix}m_1 = 0 & m_2 = -2.64575 & m_3 = 2.64575 \\[1 em] m_4 = 1.58114i & m_5 = -1.58114i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (2m^4+5m^2)(m^3-7m) &= 0&& \text{simplify left side} \\[1 em]2m^7-14m^5+5m^5-35m^3 &= 0&& \\[1 em]2m^7-9m^5-35m^3 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 2x^{7}-9x^{5}-35x^{3} = 0 } $, first we need to factor our $ x^3 $.
$$ 2x^{7}-9x^{5}-35x^{3} = x^3 \left( 2x^{4}-9x^{2}-35 \right) $$$ x = 0 $ is a root of multiplicity $ 3 $.
The remaining roots can be found by solving equation $ 2x^{4}-9x^{2}-35 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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