It is given that one root of $2x^2 + rx + s = 0$, with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$. The value of $s$ is: 
$\textbf{(A)}\ \text{undetermined}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ -13\qquad\textbf{(E)}\ 26$
If a quadratic with real coefficients has two non-real roots, the two roots must be complex conjugates of one another.
This means the other root of the given quadratic is $\overline{3+2i}=3-2i$. 
Now Vieta's formulas say that $s/2$ is equal to the product of the two roots, so
$s = 2(3+2i)(3-2i) = \boxed{26 \textbf{ (E)}}$.