What is the algebra to rearrange the constant acceleration formula? I'm working through The Cartoon Guide to Physics and I can't figure out how an equation was rearranged for the constant acceleration formula $d = {1\over2}gt^2$.  The object will fall 4 ft and the book gave $t = {\sqrt{1 sec^2/4}} = {1\over 2} sec$  as the solution.
My attempt was (edited thanks to  N. F. Taussig):
$$\begin{align}
4ft &= {1\over 2}(32ft/sec^2)t^2 \\
{4ft \over t^2} &= {1\over 2}(32ft/sec^2) \\
{1\over t^2} &= {{{1\over 2}(32ft/sec^2)} \over 4ft} \\
{1\over t^2} &= {{16ft/sec^2} \over 4ft} \\
{1\over t^2} &= {{4 sec^2 }} \\
t^2 &= {1 \over {4 sec^2}} \\
\sqrt{t^2} &= {\sqrt{1} \over \sqrt{4 sec^2}} \\
t &= { 1 \over 2 }sec
\end{align}$$
I don't see where their $t = {\sqrt{1 sec^2/4}}$ came from.  Hopefully it's not too much to ask not only where I'm going wrong but what algebra rules I need to study to make this come more naturally in the future. 

You know that 
$$d = \frac{1}{2}gt^2$$
Solve for $t$.
\begin{align*}
d & = \frac{1}{2}gt^2\\
2d & = gt^2\\
\frac{2d}{g} & = t^2\\
\sqrt{\frac{2d}{g}} & = t
\end{align*}
Substitute for $d$ and $g$ to obtain
$$t = \sqrt{\frac{2 \cdot 4~\text{ft}}{32~\frac{\text{ft}}{\text{s}^2}}} = \sqrt{8~\text{ft} \cdot \frac{1}{32}~\frac{\text{s}^2}{\text{ft}}} = \sqrt{\frac{1}{4}~\text{s}^2} = \frac{1}{2}~\text{s}$$
