<s>[INST] 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 $x(x+1)$.
 [/INST] So we have $$\sqrt{x}=6-y$$ which we substitute into the second equation to give $$(6-y)^4+y^2=90.$$  This is equivalent to finding the roots of the polynomial $$f(y):=(6-y)^4+y^2-90.$$  The Fundamental Theorem of Algebra implies this has exactly $4$ (not necessarily distinct) roots.
You mentioned one root, $3$, in your comment.  We can find the other roots comptuationally (WolframAlpha); there's closed formulas for them:
$$7+\frac{\sqrt[3]{45+\sqrt{3054}}}{3^{2/3}}-\frac{7}{\sqrt[3]{3(45+\sqrt{3054})}}$$
$$7-\frac{(1+i\sqrt{3})\sqrt[3]{45+\sqrt{3054}}}{2 \times 3^{2/3}}-\frac{7(1-i\sqrt{3})}{2\sqrt[3]{3(45+\sqrt{3054}})}$$
$$7-\frac{(1-i\sqrt{3})\sqrt[3]{45+\sqrt{3054}}}{2 \times 3^{2/3}}-\frac{7(1+i\sqrt{3})}{2\sqrt[3]{3(45+\sqrt{3054}})}$$
Then we substitute them into $\sqrt{x}=6-y$ to get the $x$-values.
The real root listed above implies $\sqrt{x} \approx 6-8.1887$ which has no solution.  Thus the solutions are $(9,3)$ and two other complex roots.
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