I was doing some problems last night, and I came across this one and I am not sure how to get started.

If the polynomial $\displaystyle Eq. A: ax^2+bx+c$ does not have any (real*) roots, prove that $\displaystyle Eq. B: ax^2+bx-c$ has two (real*) roots.

*In full disclosure, the text did not include the word 'real' before the word 'roots'. Obviously, by the Fundamental Theorem of Algebra, both will have two roots.

I was thinking about arguing that equation A must have a discriminant that is negative ($\displaystyle b^2-4ac < 0 $) and therefore Equation B will have a positive discriminant ($\displaystyle b^2-4a(-c) = b^2+4ac > 0 $). A quadratic with a positive discriminant has two real roots.

Is there a better / more clever way to solve this?

Thanks