Answer
\begin{align*}
x_1 & = 2.5 \\
x_2& = 2.45\\
x_3& = 2.44948
\end{align*}
Work Step by Step
Given $$ f(x)=x^{2}-6, \quad x_{0}=2$$
Since
$$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
Then
\begin{align*}
x_{n+1}&= x_n -\frac{x_n^{2}-6}{2x_n}
\end{align*}
Hence
\begin{align*}
x_1 &= x_0 -\frac{x_0^{2}-6}{2x_0}=2-\frac{4-6}{4}= 2.5 \\
x_2&= x_1 -\frac{x_1^{2}-6}{2x_1}=2.5-\frac{(2.5)^2-6}{2(2.5)}= 2.45\\
x_3&= x_2 -\frac{x_2^{2}-6}{2x_2}=2.45-\frac{(2.45)^2-6}{2(2.45)}= 2.44948
\end{align*}
