jmc

To find the standard form for the equation of the hyperbola, we move all the terms to one side and then complete the square in both variables: $$\begin{array}{rl}{x}^2-10x-4y^2+5& =0\\ (x^2-10x+25)-4y^2+5& =25\\ (x-5{)}^2-4y^2& =20\\ \frac{(x-5{)}^2}{20}-\frac{y^2}{5}& =1.\end{array}$$This fits the standard form of the hyperbola $$\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1,$$where $a=2\sqrt{5},$ $b=\sqrt{5},$ $h=5,$ and $k=0.$ Thus, the center of the hyperbola is the point $(h,k)=(5,0).$ Because the $x^2$ coefficient is positive and the $y^2$ coefficient is negative, the foci are horizontally aligned with the center of the hyperbola. We have $$c=\sqrt{a^2+b^2}=\sqrt{20+5}=5,$$which is the distance from the center of the hyperbola to each focus. Therefore, the two foci of the hyperbola are $(5\pm 5,0),$ which gives two points: $\boxed{(10,0)}$ and $\boxed{(0,0)}.$ (Either point is an acceptable answer.)