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 $2x^2-5x+6=0$:

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

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

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

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

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