Answer
foci $(2\pm\sqrt {41},-3)$, asymptotes $y=\pm\frac{5}{4}(x-2)-3$; see graph.
Work Step by Step
Step 1. Rewrite the equation as
$25(x^2-4x+4)-16(y^2+6y+9)=444+100-144$
or
$25(x-2)^2-16(y+3)^2=400$
and in standard form as
$\frac{(x-2)^2}{16}-\frac{(y+3)^2}{25}=1$
Step 2. We can identify $a=4, b=5, c=\sqrt {a^2+b^2}=\sqrt {41}$, center $(2,-3)$ and a horizontal transverse axis.
Step 3. We can locate the foci at $(2\pm\sqrt {41},-3)$ and asymptotes as $y+3=\pm\frac{b}{a}(x-2)$ or $y=\pm\frac{5}{4}(x-2)-3$
Step 4. We can graph the equation as shown in the figure.
