Solving an equation with first degree polynomial and an exponential For my Math C assignment I have to solve this function in regards to a falling mass with resistance 'kv' where $k=0.005$ and $g=9.8$. I have anti-derived the function from an acceleration to a velocity to finally a displacement. I would like to know how I can solve this algebraically without using graphing software, I have solved to 27.0811.
$$2400=196x+3920e^{-0.05x}-3920$$
 A: As mentioned, there is the Lambert W method.  The Lambert W function is defined so that $y=x e^x$ is equivalent to $x = W(y)$.
So in our problem
$$
2400=196x+3920e^{-x/20}-3920
$$
rearrange to get
$$
\frac{79}{49} - \frac{x}{20} = e^{-x/20}
\\
e^{79/49}\left(\frac{79}{49} - \frac{x}{20}\right) = e^{79/49 -x/20}
$$
So that if $y = \frac{x}{20} - \frac{79}{49}$ our equation becomes
$$
-e^{79/49} y = e^{-y}
\\
ye^y = -e^{-79/49}
\\
y = W\left(-e^{-79/49}\right)
\\
\frac{x}{20} - \frac{79}{49} = W\left(-e^{-79/49}\right)
\\
x = 20 \;W\left(-e^{-79/49}\right) + \frac{1580}{49}
$$
In fact $W$ is multi-valued, and we get all complex solutions of your equation by taking all the branches of $W$.  The two real solutions are
$$
20 \;W_0\left(-e^{-79/49}\right) + \frac{1580}{49} \approx 27.081065
\\
20 \;W_{-1}\left(-e^{-79/49}\right) + \frac{1580}{49} \approx -18.700406
$$
A: Your equation can be written as the equation verified by the abscissas of the intersection points of the 2 curves with equations:
$$y=3920\exp(-0.05x) \ \ \ \text{and} \ \ \ y=-196x+6320.$$
(see figure below).
In fact, there are two roots, the one you have already found, and a negative one around $-18.7$, maybe without physical significance.
It is easy to device a fixed point iteration method for finding an accurate value of the positive root ; we rewrite the equation as
$$x=f(x) \ \ \ \text{with} \ \ \ f(x):=(6320-3920\exp(-0.05x))/196$$
and build recurrence relationship 
$$x_{n+1}=(6320-3920\exp(-0.05x_n))/196 \ \ \ \text{with} \ \ \ x_0=20.$$
which converges to $L=27.081065708166108\cdots$ (the convergence is ensured by the fact that $|f'(x)|=\tfrac{39.2}{196}exp(-0.05x) < 1$ for all $x>0$.) 

In black, the curve with equation $y=3920\exp(-0.05x)$; in red, the straight line with equation $y=-196x+6320.$
Using a similar method, the other root is found to be $-18.700406404095428\cdots$.
