quick discriminant question

Just because $b^2 - 4ac = 0$ doesn't mean that there are no solutions. You could still have a solution at $\frac{-b}{2a}$.

There are no (real) solutions when $b^2 - 4ac > 0$

ooops

I know that if:
$b^2-4ac = 0$ then there are no points of intersection. Is that how you'd write it though like after substituting the values, would you leave the = 0 at the end? You don't change it to something like less than or equal to or anything do you? I mean, I don't know why you would but just to be sure.

Thanks!

I know that if:
$b^2-4ac = 0$ then there are no points of intersection.

Thanks!