Answer
Both equations have two distinct real solutions.
The solutions of one equation are the negatives of the solutions of the other.
Work Step by Step
With the assumption made by the problem text, $ax^{2}+bx+c=0$
has two real solutions , $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ .
Name them
$x_{1}=\displaystyle \frac{-b-\sqrt{b^{2}-4ac}}{2a}$ and
$x_{2}=\displaystyle \frac{-b+\sqrt{b^{2}-4ac}}{2a}$
Also, by the assumption, and because $(-b)^{2}=b^{2}$, the equation
$ax^{2}-bx+c=0$ has two solutions.
$x=\displaystyle \frac{-(-b)\pm\sqrt{(-b)^{2}-4ac}}{2a}=\frac{b\pm\sqrt{b^{2}-4ac}}{2a}$
Let $x_{3}=\displaystyle \frac{b-\sqrt{b^{2}-4ac}}{2a}$
Then , $x_{3}=-\displaystyle \frac{-b+\sqrt{b^{2}-4ac}}{2a}=-x_{2}$
And let $x_{4} =\displaystyle \frac{b+\sqrt{b^{2}-4ac}}{2a}$.
Then, $x_{4}=-\displaystyle \frac{-b-\sqrt{b^{2}-4ac}}{2a}=-x_{1}$
Both equations have two distinct real solutions.
The solutions of one equation are the negatives of the solutions of the other.
