Simple algebra simplification I have 
$ \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}$
Which simplified to 
$ -4rS^{2} \pm \sqrt{16r^{2}S^{4} +1}$
But I can’t seem to get this
 A: Note that
$$
64r^2S^4+4=4(16r^2S^4+1)
$$
so
$$
\sqrt{64r^2S^4+4}=\sqrt{4(16r^2S^4+1)}=2\sqrt{16r^2S^4+1}
$$
Finally,
\begin{align}
\frac{-8rS^2\pm\sqrt{64r^2S^4+4}}{2}
&=\frac{-8rS^2\pm2\sqrt{16r^2S^4+1}}{2}\\[6px]
&=\frac{2(-4rS^2\pm\sqrt{16r^2S^4+1}\,)}{2}\\[6px]
&=-4rS^2\pm\sqrt{16r^2S^4+1}
\end{align}

Your original equation apparently was
$$
x^2+8rS^2x-1=0
$$
Whenever you have a factor $2$ that can be extracted from the coefficient of the degree $1$ term, the simplification above can be performed: if the equation is $ax^2+2\beta x+c=0$, the quadratic formula yields
\begin{align}
\frac{-2\beta\pm\sqrt{4\beta^2-4ac}}{2a}
&=\frac{-2\beta\pm\sqrt{4(\beta^2-ac)}\,}{2a}\\[6px]
&=\frac{-2\beta\pm2\sqrt{\beta^2-ac}}{2a}\\[6px]
&=\frac{-\beta\pm\sqrt{\beta^2-ac}}{a}
\end{align}
In your case $a=1$, $\beta=-4rS^2$ and $c=-1$.
It's not necessary to remember one more formula, just to recall that the simplification can be done.
A: We have $\frac{-8rS^2}{2}=-4rS^2$ and $\frac{\sqrt{4X}}{2}=\sqrt{X}$.
A: Let $$D = \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}$$
Now
\begin{align}
D &= \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}\\
&=\frac{-8rS^{2}}{2} \pm \frac{\sqrt{64r^{2}S^{4} + 4}}{2} \\
&= -4rS^2 \pm \frac{\sqrt{64r^{2}S^{4} + 4}}{\sqrt{4}} \\
&= -4rS^2 \pm \sqrt{\frac{64r^{2}S^{4} + 4}{4}}\\
&=-4rS^2 \pm \sqrt{16r^{2}S^{4} + 1}
\end{align}
