Determining distance across face of bars following a curve So I'm not much of a mathematician, and I've been trying to figure out how one would solve this real world problem I have.
I have 2"x4" wood cut into squares (2x4x4).
I am trying to figure out at what distance to place each board on top of each other so that the top edges would follow the curve (-x^2)+25 from -5 to 0 where each increment of 1 is 1 foot.
I hope this makes sense, please let me know if i need to clarify.
 A: It's a bit easier to solve this problem by flipping the curve on its side; i.e., instead of looking at $y=-x^2+25$:
$\hskip2.7in $  
looking at $y=\sqrt{25-x}$:   
$\hskip1.4in $ 
Using the latter setup, what you want to find first (if I have understood your question correctly) are the non-negative values of $x$ where $$y=\sqrt{25-x}=\frac{n}{3}$$
where $n$ is an integer (all my numbers will be in feet, hence $4$ inches is $\frac{1}{3}$). These are just the numbers
$$25-\frac{n^2}{9}$$
as $n$ ranges from $1$ to $15$, i.e.
224/9, 221/9, 24, 209/9, 200/9, 21, 176/9, 161/9, 16, 125/9, 104/9, 9, 56/9, 29/9, 0

or, to use approximate values,
24.89, 24.56, 24.00, 23.22, 22.22, 21.00, 19.56, 17.89, 16.00, 13.89, 11.56, 9.000, 6.222, 3.222, 0

This shows how the number of wood pieces you can fit within the boundary increases by one at each of these values of $x$:

As you can see, for $x\leq25-\frac{n^2}{9}$, we can fit in $n$ wood pieces.
Now, the thickness of each wood piece is 2 inches, or $\frac{1}{6}$, so the top of the $m$th row (counting the row whose bottom is the $y$-axis as row 1) is at $x=\frac{m}{6}$. Thus, you will be able to fit $n$ wood pieces in on the $m$th row  if and only if
$$\frac{m}{6}\leq25-\frac{n^2}{9},$$
or
$$n\leq \sqrt{225-\frac{3m}{2}},$$
so the number of wood pieces you can fit in on the $m$th row is
$$\left\lfloor\sqrt{225-\frac{3m}{2}}\right\rfloor.$$
Using this formula, we can generate a side-on view of what the final result of your project will look like:

Is this what you had in mind?

Mathematica code for that final image:
f[m_] := Floor[Sqrt[225 - (3 m/2)]]

Show[ParametricPlot[Table[{m/6, u*f[m]/3}, {m, 1, 150}], {u, 0, 1}], 
Plot[Sqrt[25 - x], {x, 0, 25}, AspectRatio -> 1/5], 
ParametricPlot[Table[{u Floor[150 - 2 n^2/3]/6, n/3}, {n, 1, 15}], {u, 0, 1}], 
PlotRange -> {0, 5}]

A: You can make a table using your equation:
$\begin {array} {c c c}x&h&boards\\
-5 &50&25\\
-4& 41&20.5\\
-3&34&17\\
-2&29&14.5\\
-1&26&13\\
0 &25&12.5 \end {array}$
Then at each $x$ position pile up enough boards to get the height you want:  $25$ of them at $5$ feet.  As your boards are only $1/3$ foot in size, maybe you need more lines in the table.  Is this what you were looking for?
