$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ for Newton-Raphson method (See the link below)
$f(x) = 0.75 x^3 -2x^2-2x + 4$
$ \implies f(2) = -2$
$f'(x) = 2.25x^2 - 4x -2$
$\implies f'(2) = 9 - 8 - 2 = -1$
So, $x_1 = 2 - \frac{-2}{-1} =2-2 = 0$
$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0 - \frac{4}{-2} = 2$
Since $x_2 = x_0$, we will get $x_3 = x_1 = 0$.
So, $x_3 = 0$, and the method never converges. A choice.
https://www.math.ubc.ca/~anstee/math104/104newtonmethod.pdf