<s>[INST] How to solve equation? How to solve this equation in the set of real numbers?
$$(x^{2}+3y^{2}-7)^{2} + \sqrt{3-xy-y^2}=0$$
I tried to solve $x^{2}+3y^{2}-7=0$ and $\sqrt{3-xy-y^2}$=0 for x. But it did not help.
 [/INST] $(x^2 + 3y^2 - 7)^2=0$ and $3-xy-y^2=0$
$x^2 + 3y^2 - 7=0$ and $3-xy-y^2=0$
$x=\pm (7-3y^2)^2$... So you have two cases:
You put $x=(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve 
You put $x=-(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve . . .
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