Answer
$\left\{-\sqrt[3]{5},-\dfrac{\sqrt[3]{4}}{2}\right\}$
Work Step by Step
Using factoring of trinomials, the given equation, $
2m^6+11m^3+5=0
,$ is equivalent to \begin{align*} (m^3+5)(2m^3+1)&=0 .\end{align*} Equating each factor to zero (Zero Product Property) and solving for the variable, then \begin{array}{l|r} m^3+5=0 & 2m^3+1=0 \\ m^3=-5 & 2m^3=-1 \\\\ & m^3=-\dfrac{1}{2} .\end{array} Taking the cube root of both sides and then rationalizing the denominator (when necessary), the equations above are equivalent to \begin{array}{l|r}
m=\sqrt[3]{-5} & m=\sqrt[3]{-\dfrac{1}{2}} \\\\
m=\sqrt[3]{-1\cdot5}& m=\sqrt[3]{-\dfrac{1}{2}\cdot\dfrac{4}{4}}
\\\\
m=\sqrt[3]{-1}\cdot\sqrt[3]{5}& m=\sqrt[3]{-\dfrac{1}{8}\cdot4}
\\\\
m=-\sqrt[3]{5}& m=-\dfrac{\sqrt[3]{4}}{2} .\end{array} Hence, the real solutions of the equation $ 2m^6+11m^3+5=0 $ is the set $
\left\{-\sqrt[3]{5},-\dfrac{\sqrt[3]{4}}{2}\right\}
$.
