"Extending" the calculation of the golden ratio using square roots (not silver-ratio) I'm looking at the following formula: 

$x =\frac{-n+\sqrt{n^{2}+4n}}{2}$

For $n=1$ this this gives $0.618...$ and then $\frac n x$ gives $1.618...$ which is $\phi$, the golden ratio. 
What particularly interests me are the following values of $n$:
$1, 2, 6, 8, 12, 14, 18, 28, 32, 36$
The values (or ratios) for $\frac n x$ then give:

1.618033989
2.732050808
6.872983346
8.898979486
12.92820323
14.93725393
18.94987437
28.96662955
32.97056275
36.97366596

I can also get to these ratios using the following function:

$\dfrac n 2 + \sqrt a \times b$

Using these values:
n=1, a=5, b=0.5
n=2, a=3, b=1
n=6, a=15, b=5
n=8, a=6, b=2
n=12, a=12, b=2
n=14, a=7, b=3
n=18, a=11, b=3
n=28, a=14, b=4
n=32, a=8, b=6
n=36, a=10, b=6

Can someone explain if I'm just doing some algebra or if there is some other pattern which explains this? 
And by 'this' I mean the correlation between: $\frac{n}{\frac{-n+\sqrt{n^{2}+4n}}{2}}$ and $\dfrac n 2 + \sqrt a \times b$)

$$\frac{n}{x}=\frac{2n}{-n+\sqrt{n^2+4n}}=\frac{2n\left(-n-\sqrt{n^2+4n}\right)}{-4n}=\frac{n}{2}+\frac{\sqrt{n^2+4n}}{2}.$$
If $n$ is even, then this is
$$\frac{n}{2}+\frac{\sqrt{4\left[(\frac{n}{2})^2+n\right]}}{2}=\frac{n}{2}+\sqrt{\left(\frac{n}{2}\right)^2+n}.$$
Now, any surd $\sqrt{m}$ can be simplified to $a\sqrt{b}$ with $b$ squarefree, so apply with $m=(\frac{n}{2})^2+n$.
