jmc

To find the standard form for the equation of the hyperbola, we complete the square in both variables: $$\begin{array}{rl}16(x^2+x)-4(y^2+5y)-85& =0\\ 16(x^2+x+\frac{1}{4})-4(y^2+5y+\frac{25}{4})-85& =4-25\\ 16(x+\frac{1}{2}{)}^2-4(y+\frac{5}{2}{)}^2& =64\\ \frac{(x+\frac{1}{2}{)}^2}{4}-\frac{(y+\frac{5}{2}{)}^2}{16}& =1.\end{array}$$Therefore, the center of the hyperbola is the point $\left(-\frac{1}{2},-\frac{5}{2}\right).$ The vertices lie to the left and right of the center, and the distance from the center to each vertex is $\sqrt{4}=2.$ Thus, the vertices have coordinates $$\left(-\frac{1}{2}\pm 2,-\frac{5}{2}\right)=\boxed{\left(\frac{3}{2},-\frac{5}{2}\right)}\text{ and }\boxed{\left(-\frac{5}{2},-\frac{5}{2}\right)}.$$(Either point is a correct answer to this problem.)