Solutions to $\sqrt{x}+y=6,x^2+y^2=90$ $$\begin{gather}
\sqrt{x}+y=6 \tag{1} \\
x^2 + y^2 = 90 \tag{2}
\end{gather}$$
WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself.  However, he himself has no idea how to solve for the two variables.  I used the method of substitution to finally obtain $x(x+1)=6(21-2y)$.  Then I figured that both sides must be non-negative, and hence the value of $y$ has to be less than $11$.  Then we test for values less than $11$ which makes $6(21-2y)$ a product of two consecutive numbers.  WE get $3$ as a value of $y$, and also $7$.  But putting $7$ in (1) does not work.
But there is a flaw in my reasoning.  When I get to the 'consecutive' part, I start assuming $x$ and $y$ are integers.  Since the person who asked me is not familiar with complex numbers,  WE can assume that the solutions are real.  However, I am interested in extending the values of $x$ and $y$ beyond the reals, if there are any.
NOTE: From (2), we get the value of $y^2$ in terms of $x$.  Then we figure out the value of $x$ in (1) and then substitute $y$ for $90-x^2$, which gives us $x(x+1)$.

Another way to show there is only one real root...
$\sqrt{x} = 6 - y \implies y \le 6$  
So we are looking for roots of $f(y) = (y-6)^4 + y^2 - 90$ where $y \le 6$.  We already have $y=3$ as a root from observation.
Using $f(y) = (y-3)(y^3 -21y^2 +154y - 402)$, we need to check if the cubic factor has any  real roots in the interval.  We note, 
If $g(y) =  y^3 -21y^2 +154y - 402, g(6) = -18 < 0$ and $g'(y) = 3y^2 - 42y + 154 > 0$ so the cubic has no root with $y < 6$.  Hence $(x, y) = (9, 3)$ is the only real solution.
