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.