You can ask an LLM to explain it. But this is a neat way of calculating a root. Probably better to use the binary tree solution but this one performs better O(log log n)

class Solution {
public:
    int mySqrt(int x) {
        if (x == 0) return 0;

        int r = x;
        while (r > x / r) {   // instead of r*r > x
            r = (r + x / r) / 2; // this is Newton's method.
        }
        return (int)r; // floor(sqrt(x))
    }
};

Note that we should avoid a possible integer overflow from r*r by re-arranging the equality or using a bigger container.