Solving equations of the form $ae^x + bx +c =0$ In a recent piece of homework I needed to solve an equation of the form $ae^x + bx +c = 0$ where $a,b$ and $c$ are constants. I could not do it; no matter how I tried I either went in circles or hit a brick wall. Can someone demonstrate a general method for solving equations of this form? Please show every step in your answers.
 A: \begin{align}
a \exp(x)+b x+c&=0
\end{align}
Substitute $bx+c=y$: 
\begin{align}
a \exp((-c+y)/b)+y &= 0
\\
-y &= a\exp(-c/b)\exp(y/b)  
\\
-y/b\exp(-y/b) &= a/b\exp(-c/b) 
\end{align}
The last equation is in the form $u\exp(u)=w$,
which has a solution in terms of Lambert $W$ function:
\begin{align}
u&=W(w).
\end{align}
Hence
\begin{align}
-y/b&=W(a/b\exp(-c/b))
\\
y&=-b W(a/b\exp(-c/b))
\\
x&=-W(a/b\exp(-c/b))-c/b.
\end{align}
In particular, equation $88\exp(x)+12x-5=0$
with $a=88,b=12,c=-5$ has one real root,
since the argument of $W$ is positive:
\begin{align}
x&=-W_0(22/3\exp(5/12))+5/12
\approx-1.3970513.
\end{align}
Edit:
Lambert's W function is included in the Gnu Scientific Library (GSL)
and is freely available by means of many computer systems and packages
(wolframalpha, R, Asymptote, python scipy.special to name a few).
A: As said, there is an analytical solution in terms of Lambert function $$x=-W(d)-\frac{c}{b}$$ where $$d=\frac{a }{b}e^{-\frac{c}{b}}$$ If you do not want to (or cannot)  use Lambert function, then numerical methods are the way to go.
Probably the simplest should be Newton method which, starting from a "reasonable" guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For the example you give in comments $$f(x)=5-12x - 88e^x$$ $$f'(x)=-12-88e^x$$ A look at the graph of the function shows that the solution is close to $-1.5$. So, let start iterating at $x_0=-1.5$; the method will generates the following iterates : $-1.393646353$, $-1.397047564$, $-1.397051301$ which is the solution for ten significant figures.
