Inverting the cumulative probability function to find roots of stochastic function Given a function: 
$$f[x]=a\, \Phi \left[-x+\sigma \sqrt{\tau}\right]-\left(b+c\, e^{-d \tau}\right)\Phi \left[-x\right]$$
where $\Phi$ is the cumulative density function of the standard normal distribution:
$$\Phi\left[z\right] = \frac{1}{\sqrt{2 \pi}}\int^{z}_{-\infty}e^{\frac{-u^2}{2}} \, \mathbb{d}u $$
...how can I find $x$ which which satisifies the conditon $f[x]=0$? 
Suppose that a, b, c, $\sigma$, t are known quantities.
I am stuck trying to use inverse identities since approximations to inverse functions seem to not hold when inverting a probability function multiplied by a constant.
Also, although there is an algorithm to find $x$ through recursion:
$$x \to-\Phi^{-1}[\frac{a}{b+c\, e^{-d \tau}}\, {\Phi\left[-x+\sigma \sqrt{\tau}\right]}] \,\,\, \forall \,\,\, \tau \in \, [t,T]$$
where $\Phi^{-1}$ is the probit function.
...this does not satisfy a closed form requirement.
Acceptable answers may include closed form solutions as well as numerical approximations provided that approximations converge $ \forall_{\left|x\right|\lt ~5} \in \mathbb{R}$. I also appreciate any direction or references.
Update:
Following Synia's comments, it is apparent that $g'[x] = 0$ has at least one solution (with $ g(x) = a \, \Phi [-x+s] - \Phi[-x]$), i.e.  
$$g'[x] = -a\, \Phi'[-x+s] + \Phi'[-x] = \frac{1}{\sqrt{2\pi}} \left( -a e^{(x-s)^2/2 } + e^{ x^2/2 } \right) $$
vanishes in 
$$ x^* : (a, s) \mapsto \frac{\frac{s^2}{2}-\ln\left[a\right]}{s} \quad \|   \, a>0 $$
Thus, if $ g[x^*(a, s)] > 0 $, there is no zero, if $ g[x^*(a, s)] = 0  $, this is the only zero, and if $ g[x^*(a, s)] < 0 $, there are two zeros, hence, one has to choose which one to approximate.
 A: Since you are considering a numerical approach: If you have access to Mathematica, you can use the "Reduce" functionality to find all roots in a given interval.
In[1]:= f[x_] := (a*Erf[-Infinity, (-x + sigma*Sqrt[tau])*Sqrt[2]] 
       - (b + c*Exp[-d*tau])*Erf[-Infinity, -x*Sqrt[2]])/2
In[2]:= a = 1; b = 1; c = 1; d = 2.5; sigma = 4; tau = 2.5;
In[3]:= Reduce[f[x] == 0 && -5 < x < 5]
Out[3]: x == -1.44496

I checked that it also works on Wolfram Alpha (so a Mathematica license is not really needed).
This documentation explains how Reduce tries to find all roots:

Given rough locations for roots, we use Newton-like iterative methods
  to home in on the roots. And finally, we use Mathematica‘s interval
  arithmetic to prove that there can only be one root inside each small
  region we’ve identified. There is some subtlety here. Proving zero
  equivalence is in general undecidable—and the way this shows up is
  that in pathological cases it can in principle require unboundedly
  much precision to distinguish multiple roots from closely spaced
  single roots.

