Would use the algorithm for solving square root.
For square root, √n
x₁ = 0.5(x₀ + n/x₀)
(This formula is known and for square root, and can be derived using Newton-Raphson's approximation equation)
Where x₀ is the initial guess. x₁ becomes the new guess.
For √100.6 let our initial guess be 10, x₀ = 10, n = 100.6
Our approximation shall be to 3 decimal places. Once we get the same answer twice we stop the algorithm.
x₀ = 10, x₁ = 0.5(x₀ + n/x₀), x₁ = 0.5(10 + 100.6/10) = 10.030, x₁ = 10.030
x₂ = 0.5(x₁ + n/x₁), x = 0.5(10.030 + 100.6/10.030) ≈10.015, x₂ ≈ 10.030 (to 3 decimal places)
Since x₂≈ x₁, the algorithm stops.
So the √100.6 is ≈ 10.030 to 3 decimal places.
I hope this helps.
