Solving for $y$, given $x = \frac{-1 + \sqrt{1+ (4y/50)} }{2}$ I have the following quadratic equation:
$$x = \frac{-1 + \sqrt{1+  (4y/50)} }{2}$$
in this case $y$ is a known variable so I can solve the equation like this for $y = 600$
$$x = \frac{-1 + \sqrt{1+  (4\times 600/50)} }{2}$$
and I was looking for help on how to refactor it so I can solve it while knowing $x$ instead of $y$, but I have no idea how to do it.
I would be looking for something like $y = x$???
Thanks!
 A: It's simple algebra.
$$y=\frac{50(2x+1)^2-50}{4}=50x(x+1)$$
A: You probably know the quadratic formula: $x_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
This is the formula for the quadratic function $f=ax^2+bx+c$. In this case we have $a=1$ and $b=1$. Comparing $4ac=4c$ with $-\frac{4y}{50}$ gives  $c=-\frac{y}{50}$. Thus $f=x^2+x-\frac{y}{50}=0$.
Finally you can solve for y.
A: First see how the right hand expression is built, starting from $y$, using simple arithmetic operations
$$y\xrightarrow[\times \tfrac{4}{50}]{}\frac{4y}{50}\xrightarrow[+1]{}1+\frac{4y}{50}\xrightarrow[\sqrt{{}}]{}\sqrt{1+\frac{4y}{50}}\xrightarrow[-1]{}-1+\sqrt{1+\frac{4y}{50}}\xrightarrow[\div 2]{}\tfrac{-1+\sqrt{1+\frac{4y}{50}}}{2}=x$$
Now, starting from $x$, undo each of these 5 operations to get $y$
$$x\xrightarrow[\times 2]{{}}2x\xrightarrow[+1]{{}}2x+1\xrightarrow[{{()}^{2}}]{{}}{{(2x+1)}^{2}}\xrightarrow[-1]{{}}{{(2x+1)}^{2}}-1\xrightarrow[\times \tfrac{50}{4}]{{}}\frac{50}{4}\left[ {{(2x+1)}^{2}}-1 \right]=y $$
