Answer
$${x_{n + 1}} = \frac{{x_n^2 + 5}}{{2{x_n}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^2} - 5 \cr
& {\text{Differentiate the given function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2} - 5} \right] \cr
& f'\left( x \right) = 2x \cr
& {\text{Use the procedure Newton's Method for Approximations}} \cr
& {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 311}}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr
& {\text{form}} \cr
& {x_{n + 1}} = {x_n} - \frac{{{{\left( {{x_n}} \right)}^2} - 5}}{{2\left( {{x_n}} \right)}} \cr
& {\text{Simplifying}} \cr
& {x_{n + 1}} = \frac{{2{{\left( {{x_n}} \right)}^2} - {{\left( {{x_n}} \right)}^2} + 5}}{{2{x_n}}} \cr
& {x_{n + 1}} = \frac{{2x_n^2 - x_n^2 + 5}}{{2{x_n}}} \cr
& {x_{n + 1}} = \frac{{x_n^2 + 5}}{{2{x_n}}} \cr} $$
