The quadratic formula says that
$\displaystyle x^2 = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot 50}}{2 \cdot 1} = \frac{-9 \pm \sqrt{-119}}{2}$
So
$\displaystyle x^2 = \frac{-9 \pm i \sqrt{119}}{2}$
A brief diversion:
$\displaystyle y = -x^2 = - \left ( \frac{-9 \pm i \sqrt{119}}{2} \right ) = \frac{9 \mp i \sqrt{119}}{2}$
Again,
$\displaystyle x^2 = \frac{-9 \pm i \sqrt{119}}{2}$
So
$\displaystyle x =\pm \sqrt{\frac{-9 \pm i \sqrt{119}}{2}}$ <-- The two $\displaystyle \pm$ symbols do not have any relation to each other.
This is technically an unsimplified answer, but in your case I'd leave it like this.
So the solution set is the four points:
The two points
$\displaystyle \left ( \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right ) $
and the two points
$\displaystyle \left (- \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right ) $
-Dan
i told you, (0,0) is only a solution if you have a + (y - 5)^2. otherwise, there are NO real solutions. if you did not type the question wrong, topsquark's solutions are correct and the solutions you have are wrong. go back to the question and make sure you have the signs right