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.

Think of the standard equations of a parabola you know $y=x^2$ or $y^2=4ax$ - something squared = something linear, and the squared quantity and the linear quantity represent axes at right-angles to one another.
The vertex occurs where the squared quantity is equal to zero, ie on the axis of symmetry.
Now notice that the equation can be rewritten as $$(3x-4y)^2=5(4x+3y+12)$$
Check that the axes implied by this form are perpendicular.
