The way that cheap calculators do it is with Newton's approximation.
x=√n
x^2=n
x^2-n=0
Newton's approximation is:
x-f(x)/(dy/dx)
x-(x^2-n)/(2x) which if you include the improved approximation
(2x^2-x^2+n)/(2x)
x↔(x^2+n)/(2x)
It converges very quickly to the exact answer (if there is one) with relatively few iterations...
For example: √5 ? take an initial guess of 2 and you get:
(x^2+5)/(2x)
2, 2.25, 2.2361, 2.2360, 2.236067977499