Question 404418
It's quite clear that no {{{x >= 0}}} can satisfy the equation. ({{{e^x > 0}}}, but {{{-x <= 0}}}).  Considering the graphs of both {{{e^x}}} and -x, there is only one solution (or intersection point), and it lies between -1 and -1/2.
(This is because {{{e^(-1) - 1 < 0}}} and {{{e^(-1/2) - 1/2 > 0}}}.
Now {{{e^x  = 1 + x + x^2/2}}} for small values of x.
So to get an approximate value for the root, solve

{{{-x = 1+x + x^2/2}}}
==> {{{x^2/2 + 2x + 1  =0}}},  or 
{{{x^2 + 4x + 2 = 0}}}.
==> {{{x = (-4 +- sqrt( 16-8 ))/2 }}} 
{{{x = (-4 +- 2sqrt(2))/2 }}} 
{{{x = -2 +- sqrt(2)) }}}.  Eliminating the bottom value the answer is 
{{{x = -2 + sqrt(2)) }}}