$x=-i$ satisfies the given equation,
$\therefore x^2(x+i)-xi(x+i)+2(x+i)=0$
$\Rightarrow (x+i)(x^2-xi+2)=0$
So, $x=-i$ is one solution. Now consider,
$x^2-xi+2=0$
$x=-i$ again satisfies the above equation,
$\therefore x(x+i)-2i(x+i)=0$
$\Rightarrow (x+i)(x-2i)=0$
So, $x=-i$, $x=2i$ are two more solutions.