Find the coordinates of either of the vertices of the hyperbola \[16x^2+16x-4y^2-20y-85=0.\](Enter your answer as an ordered pair. Enter the coordinates of one of the vertices, not both.)
To find the standard form for the equation of the hyperbola, we complete the square in both variables: \[\begin{aligned} 16(x^2+x) - 4(y^2+5y) - 85  &= 0 \\ 16(x^2+x+\tfrac14)-4(y^2+5y+\tfrac{25}4) - 85 &= 4 - 25 \\ 16(x+\tfrac12)^2 - 4(y+\tfrac52)^2 &= 64 \\ \frac{(x+\tfrac12)^2}{4} - \frac{(y+\tfrac52)^2}{16} &= 1. \end{aligned}\]Therefore, the center of the hyperbola is the point $\left(-\tfrac12, -\tfrac52\right).$ 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 \[\left(-\tfrac12 \pm 2,-\tfrac52\right) = \boxed{\left(\tfrac32, -\tfrac52\right)} \text{ and } \boxed{\left(-\tfrac52, -\tfrac52\right)}.\](Either point is a correct answer to this problem.)