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3.87298335You can estimate the value to be 4 and then use Newton's method.

First approximation by

estimated guess: x1 = 4

Better approximation:

x2 = x1 - [ (x12-15) / (2x1) ]

caculated: x2 = 4 - (16-15)/(8)= 3.87(50)

Even better approximation:

x3 = x2 - [ (x22-15) / (2x2) ]

caculated: x3 = 3.875 - (3.8752-15)/(2*3.875) = 3.87(2983(87096774))

Even more better approximation:

x4 = x3 ...... etc.

caculated: x4 = 3.872983(34620745)

Each time you replace x2 with x1 and repeat the equation many times until you get two answers with enough digits that are the same.

In this example you need to repeat the procedure

  • two times to get an '2 decimal' precision answer:

    x2,round(2) = x3,round(2) = 3.87 , (between 3.875and 3.873)

    or

  • three times to get an '6 decimal' precision answer:

    x3,round(6) = x4,round(6) = 3.872983 , (between 3.8729839 and 3.8729833)

  • four times to get an '8 decimal' precision answer: x4,round(8) = 3.87298335

    (calculation not displayed)

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