Prove that quadratic equation is 0 for any integer $a, b, c$

Question

Prove that there exists a number $r$ such that $ar^2 + br + c = 0$ for any given integers $a, b, c$.

I'm stuck on this. Particularly, I see it problematic as $r$ can probably be an irrational or imaginary(?) number.

Answer 1

There are two solutions to this equation. They are$$\frac{-b+\sqrt{ b^2-4ac}}{2a}$$and $$\frac{-b-\sqrt{ b^2-4ac}}{2a}$$These root are in general complex numbers. In fact, if they have non-zero imaginary parts, then the roots are complex conjugates.