Solve $x = y^{2} - y$ for $y$

Question

Make y the subject of $x=y^2-y$

I put this into Wolfram Alpha and it said the answer was $(1\pm\frac{\sqrt{4x+1}}{2})$ but I have absolutely no idea how it got to that answer, so I'd appreciate if someone could explain it to me, thanks in advance.

Answer 1

$$x=y^2-y$$ $$y^2-y-x = 0$$ $$\color{red}{a}\cdot y^2- \color{blue}{b} \cdot y- \color{green}{c} = 0 \,\,: a=1, b=1, c =x$$

Derivation of the Quadratic equation:
\begin{array}{rcll} ay^2+by+c & = & 0 &| -c\\[1ex] ay^2+by & = & -c&|{}\cdot 4a\\[1ex] 4a^2y^2+4aby & = & -4ac&| +b^2 \text{ (completing the square )}\\[1ex] (2ay)^2+2\cdot 2ay\,b + b^2 & = & b^2-4ac &|\text{ Forming with binomic formula}\\[1ex] (2ay+b)^2 & = & b^2-4ac &| \pm \sqrt{\quad}\\[1ex] 2ay+b & = & \pm\sqrt{b^2-4ac} &|-b\\[1ex] 2ay & = & -b \pm\sqrt{b^2-4ac} &|:(2a)\\[1ex] y & = & \dfrac{-\color{blue}{b} \pm\sqrt{\color{blue}{b}^2-4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}& \end{array}

A very interesting note is: Why can ALL quadratic equations be solved by the quadratic formula?

Answer 2

This is a completing the square problem.

$$ x = y^{2} - y $$

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

$$ y = \pm \sqrt{x + \frac{1}{4}} + \frac{1}{2}$$

$$ y = \frac{1}{2} \pm \sqrt{x + \frac{1}{4}} $$

$$ y = \left[ \frac{1 \pm 2 \sqrt{x + \frac{1}{4}}} {2} \right] = \left[ \frac{1 \pm \sqrt{4x + 1}} {2} \right] $$

$$ y = \frac{1}{2} \pm \frac{\sqrt{4x + 1}}{2}$$

(I have checked my answers and I believe the Wolfram answer you quoted is incorrect. The $1$ ought to be $\frac{1}{2}.$) $$ $$