System of equations: $x^2+y=7, y^2+x=11$ 
Possible Duplicate:
Steps to solve  this system of equations 

During the flight from Moscow to Yerevan my neighbor gave me the following problem:
Solve the system:
$$\left\{\begin{array}{c}x^2+y=7 \\ y^2+x=11. \end{array}\right.$$
It is easy to find 1 of the 4 solutions. Is there a beautiful way to find the other three?
 A: This is maybe a half solution, maybe less. I tried several things and maybe these ideas inspire someone to write a complete solution. 
Note that the intersection points lie on the circle of the form
$$\left(x+\frac12\right)^2+\left(y+\frac12\right)^2=\frac{37}{2}$$
This follows from a general theorem on the intersections of two parabolas with orthogonal axes of symmetry.
Now there are so many theorems on cyclic quadrilaterals and even some statements on a parabola passing through the four corners. Then the given point $(2,3)$ should give some additional information. But I couldn't find a concise way of finishing this.
A: Every solution of the given system
$$\left\{ 
\begin{array}{c}
x^{2}+y=7 \\ 
y^{2}+x=11
\end{array}
\right. \tag{0}$$
is a solution of
$$\left\{ 
\begin{array}{c}
\left( y-3\right) \left( y^{3}+3y^{2}-13y-38\right) =0 \\ 
x^{2}=121-22y^{2}+y^{4}.
\end{array}\tag{1}
\right. $$
The same applies to the system
$$\left\{ 
\begin{array}{c}
y^{2}=49-14x^{2}+x^{4} \\ 
\left( x-2\right) \left( x^{3}+2x^{2}-10x-19\right) =0.
\end{array}\tag{2}
\right. $$
The integral solution of $(0)$ is $\left( x_{0},y_{0}\right) =\left( 2,3\right) $. Simple ways to find the remaining solutions are only possible in particular cases, as far as I know. The standard way to solve a cubic equation such as
$$y^{3}+3y^{2}-13y-38=0\tag{3}$$
is to make the change of variables $$y=s-\dfrac{3}{3\cdot 1}=s-1\tag{3a}$$ to get the reduced cubic equation
$$s^{3}-16s-23=0.\tag{4}$$
If the discriminant $q^{2}+\frac{4p^{3}}{27}$ of an equation of the form $s^3+px+q=0$ is negative, its three solutions are real numbers. In this case we have $q^{2}+\frac{4p^{3}}{27}=23^{2}-\frac{4\times 16^{3}}{27}<0$ and the solutions of $(4)$ can be written in the trigonometric form$^{1}$ 
$$s_{k}=2\sqrt{\frac{16}{3}}\cos \left( \frac{1}{3}\arccos \left( \frac{23}{2}\sqrt{\frac{27}{16^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right),\tag{5}$$
with $k=1,2,3$. So $$y_{k}=s_{k}-1\tag{6}$$ and 
$$x_{k}=11-y_{k}^2.\tag{7}$$
For $k=1$, we get $\left( x_{1},y_{1}\right) \approx \left(
-1.8479,3.5844\right) $. And similarly for $k=2$ and $k=3$.
--
$^{1}$A deduction can be found in this Portuguese post of mine.
Added. If $\Delta =q^{2}+\frac{4p^{3}}{27}<0$ the three real solutions of the following reduced cubic equation
$$t^{3}+pt+q=0\tag{A}$$
are given by
$$t_{k}=2\sqrt{-\frac{p}{3}}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}\sqrt{-\frac{27}{p^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right) \tag{B},$$
with $k=1,2,3$.
PS. I do not find trigonometric functions nor radicals ugly. But this is just an opinion. 
