To find √S: guess x, and if x is too big then S/x is too small — so average them and repeat: x → (x + S/x)/2. The true root sits exactly between a guess and its quotient, so the average always improves, converging on the fixed point where x = S/x. It is on Babylonian clay tablets (YBC 7289 gives √2 = 1;24,51,10 — three sexagesimal places, about six decimal digits — c. 1800 BCE) — the oldest numerical algorithm we have — and it is exactly Newton's method (dart 359) on x²−S, three and a half thousand years early.
The demo runs the Babylonian average for √2 from a crude guess — the fixed point where x = 2/x: live demo
“Square roots need a modern algorithm.” — a clay tablet had √2 to about six decimal digits by averaging a guess with its quotient. The oldest algorithm is a fixed-point iteration. cited
A guess and its quotient straddle the root; their average closes in. The oldest algorithm, still a fixed point. Babylon / Heron
On the canonical compiler, the Babylonian average from 1 reaches √2 = 1.4142135… — the same fixed point as dart 358, four thousand years older: