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?

$$ \sqrt{4-3y-y^2} = x(y+4)$$
$$  {4-3y-y^2} = x^2(y+4)^2$$
$${ y^2+3y-4} = -x^2(y+4)^2$$
$$(y+4)(y-1) = -x^2(y+4)^2$$
$$\frac{y-1}{y+4} = -x^2$$
$$\frac{1-y}{y+4} =  x^2$$
$$\sqrt{\frac{1-y}{y+4}} = x$$
$$ \frac{1-y}{y+4}  = x^2$$
$$1-y = x^2y + 4x^2$$
$$1-y  = x^2y + 4x^2$$
$$-y = x^2y + 4x^2-1$$
$$y = 1-x^2y - 4x^2 $$
$$y + x^2y = 1-4x^2$$
$$y(1+x^2) = 1-4x^2$$
$$y = \frac{1-4x^2}{1+x^2}$$
