Answer
$$ - 0.79$$
Work Step by Step
$$\eqalign{
& {x^2}{e^{ - x}} + {x^2} - 2 = 0;\,\,\,\,{\text{in the interval }}\left[ { - 3,0} \right] \cr
& {\text{Let }}f\left( x \right) = {x^2}{e^{ - x}} + {x^2} - 2,{\text{ so that}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{x^2}{e^{ - x}} + {x^2} - 2} \right) \cr
& f'\left( x \right) = {x^2}\left( { - {e^{ - x}}} \right) + {e^{ - x}}\left( {2x} \right) + 2x \cr
& f'\left( x \right) = - {x^2}{e^{ - x}} + 2x{e^{ - x}} + 2x \cr
& {\text{Interval }}\left[ { - 3,0} \right]{\text{ then }}a = - 3{\text{ and }}b = 0.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr
& f\left( { - 3} \right) = {\left( { - 3} \right)^2}{e^{ - \left( { - 3} \right)}} + {\left( { - 3} \right)^2} - 2 \approx 187.76 > 0 \cr
& f\left( 0 \right) = {\left( 0 \right)^2}{e^{ - \left( 0 \right)}} + {\left( 0 \right)^2} - 2 = - 2 < 0 \cr
& f\left( { - 3} \right){\text{ and }}f\left( 0 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr
& {\text{There is a solution for the equation in the interval }}\left( { - 3,0} \right) \cr
& {\text{As an initial guess }}{c_1} = - 3.{\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
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = - 3 - \frac{{{{\left( { - 3} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( { - 3} \right)}^2} - 2}}{{ - {{\left( { - 3} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( { - 3} \right){e^{ - \left( 0 \right)}} + 2\left( { - 3} \right)}} \approx - 2.388 \cr
& {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr
& {c_3} = - 2.388 - \frac{{{{\left( { - 2.388} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( { - 2.388} \right)}^2} - 2}}{{ - {{\left( { - 2.388} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( { - 2.388} \right){e^{ - \left( 0 \right)}} + 2\left( { - 2.388} \right)}} \approx - 1.834 \cr
& {\text{In the same way}} \cr
& {c_4} = - 1.834 - \frac{{{{\left( { - 1.834} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( { - 1.834} \right)}^2} - 2}}{{ - {{\left( { - 1.834} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( { - 1.834} \right){e^{ - \left( 0 \right)}} + 2\left( { - 1.834} \right)}} \approx - 1.363 \cr
& {c_5} = - 1.363 - \frac{{{{\left( { - 1.363} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( { - 1.363} \right)}^2} - 2}}{{ - {{\left( { - 1.363} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( { - 1.363} \right){e^{ - \left( 0 \right)}} + 2\left( { - 1.363} \right)}} \approx - 1.018 \cr
& {c_6} = - 1.018 - \frac{{{{\left( { - 1.018} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( { - 1.018} \right)}^2} - 2}}{{ - {{\left( { - 1.018} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( { - 1.018} \right){e^{ - \left( 0 \right)}} + 2\left( { - 1.018} \right)}} \approx - 0.837 \cr
& {c_7} = - 0.837 - \frac{{{{\left( { - 0.837} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( { - 0.837} \right)}^2} - 2}}{{ - {{\left( { - 0.837} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( { - 0.837} \right){e^{ - \left( 0 \right)}} + 2\left( { - 0.837} \right)}} \approx - 0.792 \cr
& {c_8} = - 0.792 - \frac{{{{\left( { - 0.792} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( { - 0.792} \right)}^2} - 2}}{{ - {{\left( { - 0.792} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( { - 0.792} \right){e^{ - \left( 0 \right)}} + 2\left( { - 0.792} \right)}} \approx - 0.7901 \cr
& \cr
& {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr
& {\text{Then}}{\text{. A solution for the given interval is }}x \approx - 0.79 \cr} $$