How to show this formula to get a square root of a number in "just few seconds" is true? – math.stackexchange.com

As presented there that would be "Newton's method." But it is older than Newton. It really is "the Babylonian" method.

The Babylonian method:

If you are looking for $\sqrt n$, take a guess and pulg it into:

$x_k+1 = \frac 12 (x + \frac nx_k)$

If you guess is close is close to $\sqrtn, \frac nx_k$ will will also nearly equal $\sqrt n$, but with the error on the other side of your target. Taking the mid-point is nearly on target.

This formula is algebraically equivalent to the one in your post.

Or:

If you know that $a$ is close to $\sqrt n$, and you want to give $a$ a small nudge ($x$).

$n = (a + x)^2 = a^2 + 2ax + x^2$

Since $x$ is small $x^2$ is very small, and we drop it.

$x = \frac n-a^22a$

and finally Newton's method:

If you want to find the zero of any function. Then find a point on the curve $(x,f(x))$, draw a line tangent to the curve, through that point until it intersects the x-axis. That point $x_n+1$

$x_k+1 = x_n - \frac f(x_k)f'(x_k)$

and in this case:
$f(x) = x^2 - 2$

A note of caution, Newton's method runs into problems if $f'(x_k)$ is near zero.

One more derivation:

if $n = a^2 + x$ then we do a binomial expansion on $(a^2+x)^\frac12$

$(a^2+x)^\frac12 = a + \frac12 a^-1 x - \frac 18 a^-3 x^2... $

Using just the first two terms, we once again have an equivalent formula.