How to find the vertex of a 'non-standard' parabola? $ 9x^2-24xy+16y^2-20x-15y-60=0 $ I have to find out the vertex of a parabola given by:
$$ 9x^2-24xy+16y^2-20x-15y-60=0 $$
I don't know what to do. I tried to bring it in the form:
$$ (x-a)^2 + (y-b)^2 = \dfrac {(lx+my+n)^2} {l^2+m^2} $$
but failed in doing so. Is there any other way to solve the problem? Or maybe you could help me bring the equation in the above form.

Hint: At the vertex of a parabola, the change of slope of tangent attain maximum or minimum, i.e. $y'''=0$.
