Vieta's for the Cubic

It is given that $1$ is one of the roots of $2x^3-5x^2+3$ . Let the other two roots be $\alpha$ and $\beta$ . Using Vieta's Formulas, we have:

$(1)(\alpha)( \beta) = \dfrac {-3}{2} \quad \Rightarrow \beta = \dfrac {-3}{2\alpha}$

We also know that:

$1+\alpha+\beta = \dfrac {5}{2}\quad \Rightarrow \alpha + \beta = \dfrac {5}{2} - 1 = \dfrac {3}{2} \quad \Rightarrow \alpha - \dfrac {3}{2\alpha} = \dfrac {3}{2}$

$2\alpha^2 - 3 = 3\alpha \quad \Rightarrow 2\alpha^2 - 3\alpha - 3 = 0$

$\Rightarrow \alpha = \dfrac {-(-3) \pm \sqrt{(-3)^2-4(2)(-3)}}{2(2)} = \dfrac {3 \pm \sqrt{33}}{4} = \dfrac {a \pm \sqrt{b}}{c}$

$\Rightarrow a+b+c = 3+33+4 = \boxed{40}$