If ab>0 and bc<0, then ax^2+bx+c=0 has two real solutions. (Prove this)

I started, but I don't know.

Let a, b, c be real.

Assume ab>0 and bc<0.

So bc<ab

So c>a.

So....

I don't know from here. It clearly involves showing that the discriminant, b^2-4ac is positive I would think. How I can go about doing so, not sure.

Thanks for the help.