$x=-i$ satisfies the given equation,

$\therefore x^2(x+i)-xi(x+i)+2(x+i)=0$

$\Rightarrow (x+i)(x^2-xi+2)=0$

Now consider, $x^2-xi+2=0$

$x=-i$ again satisfies the above equation,

$\therefore x(x+i)-2i(x+i)=0$

$\Rightarrow (x+i)(x-2i)=0$

$x=2i$ is another solution,

So, $x=-i$, $x=-i$, $x=2i$ are the three solutions.