Find the center of the hyperbola $3x^2-5y^2-6x+20y-32=0$

The equation of the given hyperbola is 

$3x^2-5y^2-6x+20y-32=0$

$\begin{array}{l}\Rightarrow 3\left(x^2-2x\right)-5\left(y^2-4y\right)=32\\ \Rightarrow 3(x-1{)}^2-5(y-2{)}^2=32+3-20=15\\ \Rightarrow \frac{3(x-1{)}^2}{15}-\frac{5(y-2{)}^2}{15}=1\\ \Rightarrow \frac{(x-1{)}^2}{5}-\frac{(y-2{)}^2}{3}=1\end{array}$