Answer
$\dfrac{-3}{2}, -1+\sqrt 2 i, -1-\sqrt 2 i$
Work Step by Step
Consider $P(x)=2x^3+7x^2+12x+9=0$
In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows:
$2x^3+7x^2+12x+9=(2x+3)(x^2+2x+3)=0$
This gives: $(2x+3)(\dfrac{1\pm\sqrt{1-4}}{2})=0$
This can be re-written as:
$(\dfrac{-2\pm\sqrt{4-12}}{2})=0$
$(2x+3)(\dfrac{-2\pm i\sqrt {8}}{2})=0 $
Thus the zeroes of $P(x)$ are: $\dfrac{-3}{2}, -1+\sqrt 2 i, -1-\sqrt 2 i$
