Any advice on reducing $(x + 1.07142)e^{-14x} = 33\times 10^{-10}$ I am working on a research project and came unstuck when I got to this point. I am generating a numerical solution for some variable $x$ but cannot seem to figure out how to reduce the expression below.
The equation is 
$$(x + 1.07142)e^{-14x} = 33\times 10^{-10}$$
I know the result ($x = 1.4615$..), but cant seem to figure out the solution to get that result. Any suggestions on how to go about this will be deeply appreciated. 
Thank you.
 A: There are many one dimensional root finding algorithms and any numerical analysis text will discuss them.  I like the discussion in Numerical Recipes with obsolete versions free on line.  Once you note that the root is somewhere between $1$ and $2$ any of the root finders should have an easy time.  If I just have one problem to do I often look for fixed point iteration.  If you can write your equation in the form $f(x)=x$ with $f(x)$ slowly varying it will converge rapidly.  Here I would write 
$$(x + 1.07142)e^{-14x} = 33\times 10^{-10}\\
x=-\frac 1{14}\ln\left(\frac{33\times 10^{-10}}{x + 1.07142}\right)$$
start with $x_0=1.5$ and iterate to convergence.  Writing this in a spreadsheet is easy with copy down for the iteration.  It converges to machine accuracy in six iterations.
A: There are two roots for
$$
 f(x) = \left( x+1.07142)e^{−14x}-33×10^{−10} \right),
$$
$$
x_{1} = -1.07142, \quad x_{2} = 1.461332134293102.
$$
There is no closed form solution.
The plot below of 
$$
\log f(x)
$$
shows the location of these roots. Outside the interval bounded by the roots the value of the function is negative, and the logarithm is complex

