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Use the Newton-Raphson method.
Let f(x) = x2 - 350
and let f'(x) = 2*x
Then start with any guess, x0.
The next estimate is x1 = x0 - f(x0)/f'(x0).
Continue iterations with xn+1 = xn - f(xn)/f'(xn).
If you start with x0 = 15 (quite a long way off, given that 152 is only 225), x2 is accurate to 6 decimal places (error = 8 in 10 million) and x3 to 13 dp (2 in 100 trillion).
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What is the square root of 350?
18.708286933869706927918743661583
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