Let $A=\left\{a\in R:\text{the equation}(1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0\right\}$ has atleast one real root. Let $A=\left\{a\in R:\text{the equation}(1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0\right\}$ has atleast one real root.Find the value of $\sum_{a\in A}a^2$.

What should i do in this question to find the possible values of $a$.I have no clue how to start with.Please help me.Thanks.

$2 a^2+x^3-6 x^2+i (2 x^3-2 x^2-4 x)+5 x = 0$
root of $2 x^3-2 x^2-4 x = 0$ are $-1, 0$ and $2$
$2 a^2+(-1)^3-6 (-1)^2+5 (-1) = 0$ imply $2a^2-12=0$ ie $a=\pm\sqrt{6}$
$2 a^2+(0)^3-6 (0)^2+5 (0) = 0$ imply $2a^2=0$ ie $a^2=0$
$2 a^2+(2)^3-6 (2)^2+5 (2) = 0$ imply $2a^2-6=0$ ie $a=\pm\sqrt{3}$
therefore $\sum a^2=18$
