finding relationship between roots of quadratic and equal area above and under x-axis 
i am trying to find the relationship between the x value (last column) and the root of the function f(x) which will result in a the exact same area under and above the x-axis. i have tried finding a factor between each 'k' in the last column but it may be to complex to prove?

It seems like some brute-force tactic may be the best here. In that case we want to find a number $k>n$ such that:
$$\begin{align*}&\left|\int_{m-\sqrt{n}}^{m+\sqrt{n}}\left((x-m)^2-n\right)dx\right|=\left|\int_{m+\sqrt{n}}^k\left((x-m)^2-n\right)dx\right|\Rightarrow\\
\Rightarrow&-\int_{m-\sqrt{n}}^{m+\sqrt{n}}\left((x-m)^2-n\right)dx=\int_{m+\sqrt{n}}^k\left((x-m)^2-n\right)dx\Rightarrow\\
\Rightarrow&\int_{m+\sqrt{n}}^k\left((x-m)^2-n\right)dx+\int_{m-\sqrt{n}}^{m+\sqrt{n}}\left((x-m)^2-n\right)dx=0\Rightarrow\\
\Rightarrow&\int_{m-\sqrt{n}}^k\left((x-m)^2-n\right)dx=0\overset{y=x-m}{\underset{dy=dx}{\Rightarrow}}\\
\Rightarrow&\int_{-\sqrt{n}}^{k-m}\left(y^2-n\right)dy=0\Rightarrow\\
\Rightarrow&\left.\frac{y^3}{3}-ny\right|_{-\sqrt{n}}^{k-m}=0\Rightarrow\\
\Rightarrow&\frac{(k-m)^3}{3}-n(k-m)-\frac{-\sqrt{n}}{3}-n\left(-\sqrt{n}\right)=0\overset{s=k-m}{\underset{a(n)=\frac{\sqrt{n}}{3}+n\sqrt{n}}{\Rightarrow}}\\
\Rightarrow&\frac{s^3}{3}-ns+a(n)=0\Rightarrow\\
\Rightarrow&s^3-3ns=-3a(n)
\end{align*}$$
Now, using Tartaglia's formula - or his poem about it! - we get that:
$$\begin{align*}s=&\sqrt[3]{\sqrt{\left(\frac{-3a(n)}{2}\right)^2+\left(\frac{-3n}{3}\right)^3}+\frac{-a(n)}{2}}-\sqrt[3]{\sqrt{\left(\frac{-3a(n)}{2}\right)^2+\left(\frac{-3n}{3}\right)^3}-\frac{-a(n)}{2}}=\\
=&\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}-\frac{a(n)}{2}}-\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}+\frac{a(n)}{2}}
\end{align*}$$
So
$$k=m+\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}-\frac{a(n)}{2}}-\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}+\frac{a(n)}{2}}$$
where
$$a(n)=\frac{\sqrt{n}}{3}+n\sqrt{n}$$
Note: What really affects $k$ is the value of $n$ since, $m$ just "transports" our entire problem right or left.
