Answer
$3, \dfrac{1-i\sqrt 5}{2}, \dfrac{1+i\sqrt 5}{2}$
Work Step by Step
Consider $P(x)=2x^3-8x^2+9x-9=0$
In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows:
$2x^3-8x^2+9x-9=(x-3)(2x^2-2x+3)=0$
This gives: $(x-3)(\dfrac{1\pm\sqrt{1-4}}{2})=0$
This can be re-written as:
$(x-3)(\dfrac{2\pm\sqrt{4-24}}{4})=0$
$(x-3)(\dfrac{2\pm 2\sqrt 5 i}{2})=0 $
Thus the zeroes of $P(x)$ are: $3, \dfrac{1-i\sqrt 5}{2}, \dfrac{1+i\sqrt 5}{2}$
