Answer
$-3,-2+2\sqrt {2}i, -2-2\sqrt {2}i$
Work Step by Step
Consider $P(x)=x^3+7x^2+18x+18=0$
In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows:
$x^3+7x^2+18x+18=(x+3)(x^2+4x+6)=0$
This gives: $(x-1)(\dfrac{-(-1)\pm\sqrt{1-4}}{2})=0$
This can be re-written as:
$(x+3)(\dfrac{-4\pm\sqrt{16-24}}{2})=0$
$(x+3)(\dfrac{-4\pm \sqrt {-8}}{2})=0 \implies (x+3)(\dfrac{-4\pm 2\sqrt {2}i}{2})=0$
so, we have $-3,\dfrac{-4 +2\sqrt {2}i}{2}, \dfrac{-4 -2\sqrt {2}i}{2}$
Thus the zeroes of $P(x)$ are: $-3,-2+2\sqrt {2}i, -2-2\sqrt {2}i$
