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?

For $-4<y\le 1, x>0$
We have $$\sqrt{(y+4)(1-y)}=x (y+4) \implies (1-y)(y+4)=x^2(y+4)^2 \implies \frac{1-y}{y+4}=x^2$$
$$\implies y(x)=\frac{1-4x^2}{1+x^2}.$$
See the graph of $y(x)$ below:

