I just figured it out! What I did was set cx+5 = 2x^2 - 5 and solved for C. Then once I had c (turns out to be 2x-10) I inserted 2 for X because they must be the same at X like you said. Wish I did this before wasting two tries on the problem and loosing points. But hey, what can you do!
Here is the function.
$f(x) = \left\{ \begin{gathered}
cx + 5,\quad x < 2 \hfill \\
2{x^2} - 5,\quad x \geqslant 2 \hfill \\
\end{gathered} \right.$ Now here is a standard notation: $\displaystyle{\lim _{x \to {c^ \pm }}}f(x ) = f(c\pm)$
Thus in this question: $f(2-)=2c+5~\&~f(2+)=3$.
In order to be everywhere continuous we must have $f(2+)=f(2-)$.