Answer
$$1.13$$
Work Step by Step
$$\eqalign{
& 5{x^2} - 3x - 3 = 0;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {1,2} \right] \cr
& {\text{let }}f\left( x \right) = 5{x^2} - 3x - 3,{\text{ so that}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {5{x^2} - 3x - 3} \right) \cr
& f'\left( x \right) = 10x - 3 \cr
& {\text{interval }}\left[ {1,2} \right]{\text{ then }}a = 1{\text{ and }}b = 2 \cr
& {\text{check }}f\left( a \right){\text{ and }}f\left( b \right) \cr
& f\left( 1 \right) = 5{\left( 1 \right)^2} - 3\left( 1 \right) - 3 = - 1 < 0 \cr
& f'\left( 2 \right) = 5{\left( 2 \right)^2} - 3\left( 2 \right) - 3 = 11 > 0 \cr
& f\left( 1 \right){\text{ and }}f\left( 2 \right){\text{ have opposite signs}}{\text{, then}} \cr
& {\text{ there is a solution for the equaion in theinterval }}\left( {1,2} \right) \cr
& {\text{As an initial guess }}{c_1} = 1 \cr
& {\text{A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr
& {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr
& {\text{then }} \cr
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{5{{\left( 1 \right)}^2} - 3\left( 1 \right) - 3}}{{10\left( 1 \right) - 3}} = \frac{8}{7} = 1.1428 \cr
& {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr
& {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = \frac{8}{7} - \frac{{5{{\left( {8/7} \right)}^2} - 3\left( {8/7} \right) - 3}}{{10\left( {8/4} \right) - 3}} = \frac{{467}}{{413}} \approx 1.1307 \cr
& {\text{In the same way}} \cr
& {c_4} = 1.1307 - \frac{{5{{\left( {1.1307} \right)}^2} - 3\left( {1.1307} \right) - 3}}{{10\left( {1.1307} \right) - 3}} = 1.1306 \cr
& {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr
& {\text{Then}} \cr
& {\text{A solution for the given interval is }}x \approx 1.13 \cr} $$
