jmc

To get the hyperbola in standard form, we complete the square in both variables: $$\begin{array}{rl}4(x^2-\frac{3}{2}x)+2& =y^2-10y\\ 4(x^2-\frac{3}{2}x+\frac{9}{16})+2+25& =(y^2-10y+25)+\frac{9}{4}\\ 4{\left(x-\frac{3}{4}\right)}^2+27& =(y-5{)}^2+\frac{9}{4}\\ \frac{99}{4}& =(y-5{)}^2-4{\left(x-\frac{3}{4}\right)}^2\\ 1& =\frac{(y-5{)}^2}{99/4}-\frac{{\left(x-\frac{3}{4}\right)}^2}{99/16}\end{array}$$It follows that the center of the hyperbola is $\boxed{\left(\frac{3}{4},5\right)}.$