<s>[INST] Solve $\begin{cases} x + y + z = 2 \\ 2xy - z^2 = 4 \\ \end{cases} $ for x, y, z. It came to my mind to rewrite the expression above as 
$$\begin{cases}
x + y = 2 - z \\ 
2xy = (2 - z)^2 + 4z \\
\end{cases}
$$
and see if there any restrictions on the values of the variables occur.
What I can see is that $(2 - z)^2 + 4z \gt 0$ for all $z$ (since $z^2 > -4$) but when I express $y$ in terms of $z$ and $x$ 
$$\begin{cases}
y = 2 - z - x\\ 
x(x - (2 - z)) = (2 - z)^2 + 4z \\
\end{cases}
$$
I see that $x(x - (2 - z)) \gt 0$ not for all $z$. This is all I got for now.
The answer is $(2, 2, -2)$.
 [/INST] You have:
$$
x + y= 2-z\\
xy = 2 + \frac{z^2}{2}.
$$
From Vieta's formulas, $x$ and $y$ are roots of
$$
t^2 - (2-z)t + 2 + \frac{z^2}{2} = 0;
$$
hence
$$
t = \frac{2 - z \pm \sqrt{(2-z)^2 - 8 - 2z^2}}{2} = \frac{2 - z \pm \sqrt{-(z+2)^2}}{2}.
$$
So, we have $z=-2$, and $t=2$; therefore $t=x=y=2$ and $z=-2$.
Greate thanks to @Shailesh for notes.
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