<s>[INST] How to solve polynomial rational relations for $y$ (e.g $\sqrt{4-3y-y^2} = x(y+4)$)? From time to time, I struggle to solve polynomial relations for $y$.
A trivial example is :
$$ \frac{y}{x} = x \iff y = x^2$$
Easy.

But consider this relation:
$$ \sqrt{4-3y-y^2} = x(y+4)$$
No matter how much I mess around it, seems impossible to bring it in $y = f(x)$ form.

*

*$  \frac{\sqrt{4-3y-y^2}}{(y+4)}= x  $

*$ 4 - 3y - y^2 = x^2(y^2 + 8y + 16) \iff (x^2-1)y^2+y(-8x-3)+4(1-4x) = 0$

Is there a trivial methodology that I am missing or is it indeed impossible to inverse some relations?
 [/INST] \begin{align*}
\sqrt{4-3y-y^2} &= x(y+4)\\
\sqrt{(1-y)(y+4)} &= x(y+4)\Rightarrow y\in[-4,1],x\in[0,+\infty)\\
\sqrt{y+4}\left(\sqrt{1-y}-x\sqrt{y+4}\right) &=0\Rightarrow \boxed{y=-4}\\
\sqrt{1-y}&=x\sqrt{y+4}\\
\sqrt{\frac{5}{y+4}-1}&=x\\
\boxed{y=\frac5{x^2+1}-4}\in(-4,1]\,\forall\,x\in[0,+\infty) 
\end{align*}
</s>