Now we can solve the quadratic equation (*)  for $x$ in terms of $y$ :
$x = \frac{-B\ y \pm \sqrt{B^2 y^2 -4 AC y^2}}{2A} = \frac{-B \pm \sqrt{B^2 -4 AC}}{2A}y$

giving ideal points : <
$-B \pm \sqrt{B^2 -4 AC},2A,0$ >

So has exactly two distinct [but not very pretty] ideal points .