What is the solution to this equation: $e^x+2x=0$ I cant find any program that actually solves this type of equations and I cant find anything helpful about this type. What is the name of these equations and how do I solve this one? Thanks.

This is a transcendental equation. As said by others, it has no closed-form expression using the standard functions.
The derivative of the function is
$$(f(x))'=(e^x+2x)'=e^x+2.$$ As it is positive, the function is increasing and has at most one root.
Now we can search for a change of sign, by trial an error.
We first observe $$f(0)=1$$ and the next value to try must be negative. Then
$$f(-1)=e^{-1}-2<0$$ and we know that the root is in $(-1,0)$.

From this, we draw a first approximation, by the secant method, giving 
$$x\approx\frac{e}{1-3e}=-0.379921\cdots$$
Now by one iteration of Newton,
$$x\approx -0.379921-\frac{e^{-0.379921}+2(-0.379921)}{e^{-0.379921}+2}\approx-0.351631$$
and so on.
