Solutions to $z^3 - z^2- z =15 $ Find in the form $a+bi$, all the solutions to the equation
$$z^3 - z^2- z =15 $$
I have no idea what to do - am I meant to factor out z to get $z(z^2-z-1)=15$ or should I plug in $a+bi$ to z? 
Please help!!!!!!!  

Real solution:
$$z^3 - z^2- z =15<=> $$
$$z^3-z^2-z-15=0<=>$$
$$(z-3)(z^2+2z+5)=0<=>$$
So:
$$z-3=0<=>z=3$$
Or:
$$z^2+2z+5=0<=>z^2+2z=-5<=>z^2+2z+1=-4<=>(z+1)^2=-4$$
($(z+1)^2=-4$ hasno solution since for all $z$ on the real line, $(z+1)^2$ is bigger or even to zero and $-4$ is smaller than zero)
Complex solution:
$$z^3 - z^2- z =15<=> $$
$$z^3 - z^2- z -15=0<=>$$
$$(z-3)(z^2+2z+5)=0<=>$$
So:
$$z-3=0<=>z=3$$
Or:
$$z^2+2z+5=0<=>z^2+2z=-5<=>z^2+2z+1=-4<=>(z+1)^2=-4<=>$$
(the square root of both sides)
$$z+1=2i<=>z=-1+2i$$
Or:
$$z+1=-2i<=>z=-1-2i$$
