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?

Hint:
$$4-3y-y^2=\dfrac{25-(2y+3)^2}4$$
needs to be perfect square of a rational number
