jmc

$$\begin{array}{ccccc}\text{multicolumn}2r{x}^2& +\left(\frac{c-7}{2}\right)x& +5& & \\ \text{cline}2-52x+7& 2x^3& +c{x}^2& -11x& +39\\ \text{multicolumn}2r-2x^3& -7x^2& & & \\ \text{cline}2-3\text{multicolumn}2r0& (c-7)x^2& -11x& & \\ \text{multicolumn}2r& -(c-7)x^2& -x(c-7)\left(\frac{7}{2}\right)& & \\ \text{cline}3-4\text{multicolumn}2r& 0& -x\left(\frac{7c-27}{2}\right)& +39& \\ \text{multicolumn}2r& & -10x& -35& \\ \text{cline}4-5\text{multicolumn}2r& & -x\left(\frac{7c-27+20}{2}\right)& 4& \end{array}$$In the last step of the division, we have $39$ left as the constant term in our dividend and we need a remainder of $4$ at the end. Since our divisor has a term of $7$, the only way to do this is if our quotient has $5$ which gives us $7\cdot 5=35$ to subtract from our dividend and get the right remainder.

Then, we need the rest of our remainder to be $0$. This means $$\frac{7c-27+20}{2}=0$$which gives us $$c=\boxed{1}.$$