You can look at this as linear extrapolation using d[x²]/dx = 2x, which is Newton's Method. Taking √p as your initial guess, your new guess is (√p + s/√p)/2 = √p + (s/√p - √p)/2 = √p + (s-p)/(2√p). You can go for another iteration if you want a really accurate square root, which roughly doubles your number of accurate digits each time: 5.75 + (33 - 5.75²)/(2·5.75) ≈ 5.744565, while the correct answer is closer to 5.74456264653802865980. At that point, though, I think it's probably easier to calculate it as (5.75 + 33/5.75)/2.
You can even use this approach for taking square roots of fairly large numbers, just using the squares of single digits — factor the initial number into a power of 100 and a number between 1 and 100. For example, 80802363 is 80.802363 × 100³, so its square root is close to 9 × 10³.
You can even use this approach for taking square roots of fairly large numbers, just using the squares of single digits — factor the initial number into a power of 100 and a number between 1 and 100. For example, 80802363 is 80.802363 × 100³, so its square root is close to 9 × 10³.
Edit: fixed stupid braino in first sentence.