# 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?

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This looks like a quadratic equation in $y$, then ... –  Claude Leibovici Apr 7 at 9:34

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.

$$y=\frac{2\pm \sqrt{4+4c}}{2}$$

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

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

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This is only completing the square to the term: $Cy^2+Dy$. You can do it by using this: $$Cy^2+Dy=C\left(y^2+\dfrac{D}{C}y\right)=C\left(y+\dfrac{D}{2C}\right)^2-C\left(\dfrac{D}{2C}\right)^2.\tag1$$ Using equation $(1)$, we complete the square the term: $y^2-2y$. We obtain \begin{align} \color{red}{y^2-2y}&=1\left(y^2+\dfrac{(-2)}{(1)}y\right)\\ &=1\left(y+\dfrac{(-2)}{2(1)}\right)^2-1\left(\dfrac{(-2)}{2(1)}\right)^2\\ &=(y-1)^2-(-1)^2\\ &=\color{blue}{(y-1)^2-1}.\tag2 \end{align} Now, plug in $(2)$ into your equation: \begin{align} \color{red}{y^2 -2y} &= x^3 + 2x^2 + 2x + 4\\ \color{blue}{(y-1)^2-1}&=x^3 + 2x^2 + 2x + 4\\ \color{blue}{(y-1)^2-1}\color{red}{+1}&=x^3 + 2x^2 + 2x + 4\color{red}{+1}\\ (y-1)^2&=x^3 + 2x^2 + 2x + 5\\ y-1&=\pm\sqrt{x^3 + 2x^2 + 2x + 5}\\ \Large\color{blue}y&=\Large\color{blue}{1\pm\sqrt{x^3 + 2x^2 + 2x + 5}}. \end{align}

$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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