Well this is simple a calculator type problem...but if you are curious as the the algorithm used by simple calculators and such...
They use a Newtonian approximation until it surpasses the precision level of the calculator or computer program..
A newtonian approximation is an interative process that gets closer and closer to the actual answer to any mathematical problem...it is of the form:
x-(f(x)/(df/dx))
In a square root problem you wish to know:
x=√n where x is the root and n is the number
x^2=n
x^2-n=0
So f(x)=x^2-n and df/dx=2x so using the definition of the newton approximation you have:
x-((x^2-n)/(2x)) which simplifies further to:
(2x^2-x^2+n)/(2x)
(x^2+n)/(2x), where you can choose any starting value of x that you desire (though convergence to an exact (if possible) solution will be swifter the closer xi is to the actual value x)
In this case the number, n=95.54, so a decent starting value for x would be 10.
Using this initial x in (x^2+95.54)/(2x) will result in the following iterative sequence of x.
