What are the ways one may solve $2x+e^x=0$? What are the ways one may solve $2x+e^x=0$ ?
Thank you.
 A: The  only analytical solution is given using Lambert function which leads $$x=-W\left(\frac{1}{2}\right)\approx -0.351734$$ 
The remaining solution involve numerical methods, such as Newton. Considering $$f(x)=2x+e^x\implies f'(x)=2+e^x$$ the fist derivative being always positive, there is only one solution. If you start using, as a guess, $x_0=0$, you will obtain the following iterates
$$x_1=-0.33333333333333333333$$ $$x_2=-0.35168933155541534177$$ $$x_3=-0.35173371099294263019$$ $$x_4=-0.35173371   124919582602$$ which is the solution for twenty significant figures.
A: The Lambert W function can be used to solve such equations:
$$x=-W(1/2)$$
A: Hint: Fixed point iteration.
$x = -\frac{1}{2}e^{-\frac{1}{2}e^{-\frac{1}{2}e^{\cdots}}}$
A: Graphically. For instance
$$
\frac{e^x}{x}=-2
$$
and plot the function $x \mapsto \frac{e^x}{x}$. The point $x=0$ is harmless, since it is not a solution of yout equation.
A: If you rewrite the equation as $e^x=-2x$, then sketch the graphs of the exponential curve $y=e^x$ and the negatively sloped line $y=-2x$, you can see there is a single point of intersection in the interval $-1\lt x\lt0$. (Actually it should be easy to see that the intersection occurs in the interval $-1/2\lt x\lt0$.)  As other answers have noted, the solution can be expressed analytically with the Lambert function, but otherwise can only be approximated numerically.
