jmc

We can factor the numerator to get $$\frac{3x^2+16x+5}{2x^2+7x-c}=\frac{(x+5)(3x+1)}{2x^2+7x-c}.$$Since there is a hole at $x=-5$ (rather than an asymptote), we must have a factor of $x+5$ in the denominator that cancels out with the corresponding factor in the numerator. So by the Factor theorem, $$2(-5{)}^2+7(-5)-c=0$$which we can solve for $c$ to get $c=15$. Then the denominator is $2x^2+7x-15$ which can be factored as $(2x-3)(x+5)$. Hence, the vertical asymptote is given by $\boxed{x=\frac{3}{2}}$.