Can any non-vertical line be translated such that it intersects a parabola exactly twice?

Question

Given the parabola $y=ax^2+bx+c$ and the line $y=mx+d$, where $(a,b,m)\not=0$ and $(a,b,c,m,d)\in\mathbb R$, does there exist an $m$ where, for all $d$, there is at most one solution to this system of equations? Or, in other words, does $$ax^2+bx+c=mx+d\implies ax^2+(b-m)x+(c-d)=0$$ have two solutions for some $d$?

Answer 1

That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?

Answer 2

The composition of 2 translations is a translation. First translate the parabola to the form $y = ax^2.$ Then you can translate the line to $y = mx + \text{sgn}(a).$