Solution:

$\Rightarrow f(x)=x^{4}-x^{3}+10 x+7=0 \\ $

The root lies between $-2$ and $-1$,

Let $x_{0}=-2$

$f^{\prime}(x)=4 x^{3}-3 x^{2}+10 \\ $

By the Newton - Raphson method,

$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \\ $

$f\left(x_{0}\right)=f(-2)=11\\ $

$f^{\prime}\left(x_{0}\right)=f^{\prime}(-2)=-34 \\ $

$\therefore x_{1}=x_{0}-\frac{f\left(x_{0}\right)}{f\left(x_{0}\right)}=-2-\frac{11}{(-34)} \\ $

$\Rightarrow x_{1}=-1.6765 \\ $

$ …