For example, suppose we want to find the zeroes of f(x) = x² – 2x – 3.  Now, we all know that you can simply factor this out to get x = 1 and x = 3, but for the sake of understanding the material, let's solve for the positive root using the Newton-Raphson Method.  The method will work with any function, regardless of capabilities of factoring or not.  

         Our function is f(x) = x² – 2x – 3 and the derivative is 2x – 2.  Let's guess an initial value of 4.  Our formula would be:

After computations, we obtain:

x₀ = 4.0000000

x₁ = 3.1666667

x₂ = 3.0064103

x₃ = 3.0004836

x₄ = 3.0000001

x₅ = 3.0000000

x₆ = 3.0000000

x₇ = 3.0000000

The calculations converge to 3.0 (which is our already established root). 

Nifty huh?