Find the zeroes of the function using Newtons method.

f(x)=(1/4)x^3-3x^2+(3/4)x-2

I find that when x=11.803 y=0

Then I do Newtons method ((xn))-((f(x)n))/((f'(x)n))

((x))-((1/4x^3-3x^2+3/4x-2))/((3/4x^2-6x+3/4))

I find that when x=11.803 y=11.803

So Did I do this correctly?