solve equation $(2x + 1)e^{-2x} = 0.3$ How can I solve the equation
$$
(2x + 1)e^{-2x} = 0.3
$$
for $x$?
I don't know how to solve this kind of equations, even don't know what kind of knowledge should I use to solve this equation.
 A: In general an equation like this one will not have a closed form solution.  I.e. you cannot solve for x symbolically.
You can certainly solve it numerically.  A good first step is to plot it the LHS using a graphing calculator.  
In this particular case, the value of the LHS is 1 when x = 0, and then declines to the left and the right (you can take the derivative to see that it has a max at 0).  So there will be two solutions, one positive and one negative.
(And by the way, while the solution can be expressed in terms of Lambert function, since that function needs to be calculated numerically, there is no real advantage relative to just solving it numerically directly -- e.g. using solver in Excel.)

A: Multiply by $-e^{-1}$ so it reads $$(-2x-1)e^{-2x-1}=-0.3e^{-1}$$ Substitute $y=-2x-1$ to get $$ye^y=-0.3e^{-1}$$ Now read about the Lambert $W$ Function to solve for $y$ (in terms of $W$). And form there, solve for $x$ in terms of $y$ (in terms of $W$).
Applying $W$ to both sides gives $y=W\mathopen{}\left(-0.3e^{-1}\right)\mathclose{}$, from which you have $$x=\frac{-1-W\mathopen{}\left(-0.3e^{-1}\right)\mathclose{}}{2}$$
There are two branches of $W$ for real applications, both with $-0.3e^{-1}$ in their domain. So there will be two solutions.
A: The Lambert $W$-function is a function $W(z)$ such that $z=W(z)e^{W(z)}$ for all $z$. There are actually different branches for $W$, so it is wrong to treat it as a single function.
Now, you are trying to solve:
$$-(2x+1)e^{-(2x+1)}=-0.3e^{-1}$$
So $2x+1=-W(-0.3e^{-1})$, so $x=\frac{-W(-0.3e^{-1})-1}{2}$. 
The Wikipedia article has ways to compute $W$. For $-e^{-1}<z<0$, as this case, there are two real values of $W(z)$.
A: As a hint
Using logarithms, we say:
$\ln[(2x + 1)e^{-2x}] = \ln(0.3)$
$\ln(e^{-2x}) + \ln(2x + 1) = \ln(0.3)$
$-2x + \ln(2x + 1) = \ln(0.3)$
$2x = \ln(2x + 1) - \ln(0.3)$
$2x = \ln\left(\dfrac{2x + 1}{0.3}\right)$
$e^{2x} = \dfrac{2x + 1}{0.3} = \dfrac{20}{3}x + \dfrac{10}{3}$
This is relatively easy to sketch without a calculator.
Having completed the sketch, you are then in a good position to apply a numerical method, as suggested by T_M in the comments and by P. Michaels in their answer.
