It is given that 2 is one root of the equation, therefore $(x-2)$ must be a factor of the expression $2x^3-9x^2+16x-12$.

And the new equation can be written as

$$ (x-2)(2x^2-5x+6)=0 $$

Then we apply the Quadratic Formula to $2x^2-5x+6=0$:

$$ x=\frac{-b \pm\sqrt{b^2 - 4ac}}{2a} $$

$$ x=\frac{5 \pm\sqrt{5^2 - 4(2)(6)}}{2\times2} $$

$$ x=\frac{5 \pm\sqrt{-23}}{4} $$

$$ x=\frac{5 \pm\sqrt{23}\sqrt{-1}}{4} $$

$$ \therefore x=\frac{5 \pm\sqrt{23}i}{4} $$