Complete the squares for x and y:
x2+8x=x2+8x+16-16=(x+4)2-16;
4y2+6y=4(y2+3y/2+9/16-9/16)=4(y+¾)2-9/4.
The equation becomes:
(x+4)2-16-4(y+¾)2+9/4-24=0. Combine the constants:
(x+4)2-4(y+¾)2=40-9/4=(160-9)/4=151/4, which is a standard hyperbola of the form:
x2/a2-y2/b2=1.
We're not asked to find a or b, just to classify the conic section.
However, it's fairly straightforward to find a and b, the centre (origin) and foci of the hyperbola, and its asymptotes now that the equation has been simplified.