Q:

Solving for $y$ in terms of $x$

I have this equation $y^2 -2y = x^3 + 2x^2 + 2x + 4$
How to solve this for $y$ in terms of $x$, because we have also the term $-2y$ on the left side? 

A:

Very quick, and dirty solution
$$y^2 -2y = x^3 + 2x^2 + 2x + 4$$
$$y^2 -2y - \left(x^3 + 2x^2 + 2x + 4\right) = 0$$
Let $c=x^3 + 2x^2 + 2x + 4$
Thus, $$y^2-2y-c=0$$
is a quadratic in $y$, apply the quadratic formula. 

$$y=\frac{2\pm \sqrt{4+4c}}{2}$$
$$y=1\pm\sqrt{c+1} = 1\pm\sqrt{x^3+2x^2+2x+5}$$

A:

$$y^2 -2y = x^3 + 2x^2 + 2x + 4 \Rightarrow $$ $$y^2-2y+1-1=x^3+2x^2+2x+4 \Rightarrow$$ $$ (y-1)^2-1=x^3+2x^2+2x+4 \Rightarrow$$ $$ (y-1)^2=x^3+2x^2+2x+5 \Rightarrow $$ $$\sqrt{(y-1)^2}=\sqrt{x^3+2x^2+2x+5} $$ 

