Different way of finding square roots of numbers? this is my first post here and I wanted to run something by people who are more knowledgeable than I. In the past couple days I've decided that I need to teach myself more advanced mathematics as it is of great importance to my studies in physics, astronomy and biology. In school we never went past algebra 1 and geometry and that has proven to be a hindrance. So I decided to brush up on the basic math that I learned in school and then move on to more advanced maths. I decided to make up a bunch of problems to solve while waiting for my books to arrive and I ran into my old nemesis, finding square roots of large numbers. I was taught prime factorization, which didn't work for this number, and the Babylonian Method, which can yield accurate results depending on the accuracy of your initial estimate and how many times you're willing to run the equation. What I wanted was an equation that used simple math and a known number instead of an assumed estimate (that must be high in the case of the Babylonian Method) that didn't have to be high or low that would yield an equally/greater accurate estimate of the square as does the BM, and would only have to be run once. I realize this is simple mathematics compared to what is normally discussed here, so I apologize, and also please forgive me if my terminology isn't quite correct. What I came up with was 

√S≈((S/2)/(e/2)+e)/2

in this S is the number we are trying to find the square of and e is the closest perfect square root regardless of weather it is high or low. 

For example, using 1,863 the number I was initially trying to find the square of, we know that the closest perfect square root is 43. So it would look like this

√1,863≈((1863/2)/(43/2)+43)/2

√1,863≈((931.5)/(21.5)+43)/2

√1,863≈(43.325+43)/2

√1,863≈ 86.325/2

√1,863≈ 43.163

Actual square of 1863=43.1624

so the estimate it yielded was very close. I've run a bunch of numbers through it both large and small and so far it appears to give better estimates than the BM (depending on the accuracy of your initial high estimate and number of times you run it through the equation) and without the requirement that the estimate is high, and also we don't have to make an initial estimate ourselves as the starting estimate is fixed for us by the nature of the equation. 

What do you guys think? Is there something I've missed or perhaps is there an easier equation like it that I am not familiar with?

Thanks!
The relation $\sqrt S\approx\dfrac{1}{2} \left(\dfrac{\frac{S}{2}}{\frac{ e}{2}}+e\right)$ can be simplified to 

$\sqrt S\approx\dfrac{1}{2} \left(\dfrac{S}{e}+e\right) $

if you square both sides you get (trust me)

$S\approx\dfrac{1}{4} \left(\dfrac{S^2}{e^2}+e^2+2S\right)$

the right side can be written as (trust me again)

$S\approx S+ \dfrac{\left(S-e^2\right)^2}{4 e^2}$

$S$ can be cancelled from both sides and we have

$\dfrac{\left(S-e^2\right)^2}{4 e^2}\approx 0$

Numerator is very very little because we choose $e $ close to $\sqrt S$ while denominator is about $4S$ therefore the fraction is very little and the approximation works

The larger the $S$ the more accurate is the approximation

I tried $\sqrt{123456789}$ and with you formula I got $11111.1110611$

while the actual value is $11111.1110605$

The only problem is to know that the  square closest to $123456789$ is $11111^2$

Hope this helps :)
If you allow yourself to use a table of square roots, the best method is to lookup the closest entry in the table and start the Babylonian method from there. It is hards to beat that (I mean in terms of choosing an initial guess).

But what you describe is strictly the Babylonian method, nothing new. Note that it is wrong that the initial estimate must be by excess.