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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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11y ago
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Q: How do you find the square root of 350?
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