Step 1:
$4x^2-5y^2-16x+10y+31=0$
Completing the squares,
$4(x^2-4x+4)-5(y^2-2y+1)=-31+16-5$
$4(x-2)^2-5(y-1)^2=-20$
The above equation is divided by $-20$ we get,
$\large\frac{-(x-2)^2}{5}+\large\frac{(y-1)^2}{4}$$=1$
Step 2:
$\large\frac{(y-1)^2}{4}+\large\frac{(x-2)^2}{5}$$=1$
The centre is at $(2,1)$ and $a^2=4,b^2=5$
$\Rightarrow a=2,b=\sqrt 5$
Let $2\alpha$ be the angle between the asymptotes.
$2\alpha=2\tan^{-1}\large\frac{b}{a}$
$\Rightarrow 2\tan^{-1}\large\frac{\sqrt 5}{2}$
The angle between the asymptotes is $2\tan^{-1}\large\frac{\sqrt 5}{2}$