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 \\ $

$ f\left(x_{1}\right)=f(-1.6765)=2.8468 \\ $

$ f^{\prime}\left(x_{1}\right)=f^{\prime}(-1 .(765)=-17.2802. \\ $

$ x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)} \\ $

$ =-1.6765-\frac{2.8468}{(-17.2802)} \\ $

$ \Rightarrow x_{2}=-1.5118 \\ $

$ \begin{aligned} &f\left(x_{2}\right)=f(-1.5118)=0.561\\\\ &f^{\prime}\left(x_{2}\right)=f^{\prime}(-1.5118)=-10.6777\\\\ &\therefore x_{3}=x_{2}-\frac{f\left(x_{2}\right)}{f\left(x_{2}\right)}\\\\ &=-1.5118-\frac{0.561}{(-10.6777)}\\\\ &\Rightarrow x_{3}=-1.4593\\\\ &f\left(x_{3}\right)=f(-1.4593)=0.0497\\\\ &f^{\prime}\left(x_{3}\right)=f^{\prime}(-1.4593)=-8.8193\\\\ &\text { Now, } x_{4}=x_{3}-\frac{f\left(x_{3}\right)}{f^{\prime}\left(x_{3}\right)}\\\\ &=-1.4593-\frac{0.0497}{(-8.8193)}\\\\ &\Rightarrow x_{4}=-1.4537\\\\ &f\left(x_{4}\right)=t(-1.4537)=0.0008\\\\ &f^{\prime}\left(x_{4}\right)=f^{\prime}(-1.4537)=-8.6278\\\\ &\therefore x_{5}=x_{4}-\frac{f\left(x_{4}\right)}{f^{\prime}\left(x_{4}\right)}\\\\ &=-1.4537-\frac{0.0008}{(-8.6278)}\\\\ &\Rightarrow x_{5}=-1.4536 \\ \end{aligned} $

Since , $X_4$ and $X_5$ are same up to three decimal places, the root is -1.453.