Let $x$ be a positive number such that $2x^2 = 4x + 9.$ If $x$ can be written in simplified form as $\dfrac{a + \sqrt{b}}{c}$ such that $a,$ $b,$ and $c$ are positive integers, what is $a + b + c$?
First, we move all terms to one side to get $2x^2 - 4x - 9 = 0.$ Seeing that factoring will not work, we apply the Quadratic Formula: \begin{align*}
x &= \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-9)}}{2 (2)}\\
&= \frac{4 \pm \sqrt{16 + 72}}{4} = \frac{4 \pm \sqrt{88}}{4}\\
&= \frac{4 \pm 2\sqrt{22}}{4} = \frac{2 \pm \sqrt{22}}{2}.
\end{align*}Since $x$ is positive, $x$ can be written as $\dfrac{2 + \sqrt{22}}{2},$ so our answer is $2 + 22 + 2 = \boxed{26}.$