How can I solve this equation (contains error function)? Edited out incorrect formula
Can someone please solve this equation for x? I have no idea what to do with the $\mathrm{erf}$ (error function). 
Edit: Hm, it did not work correctly... here is the function I meant to solve for x in symbolic form:
$$f(x) = a*\left(0.5*\mathrm{erf}\left(\frac{x-b}{c\sqrt{2}}+.5\right)\right)+d$$
Coefficients (with 99% confidence
bounds):
       a =       1.412  (1.411, 1.412)
       b =       1.259  (1.259, 1.259)
       c =       1.003  (1.002, 1.003)
       d =      0.3016  (0.3014, 0.3017)

When I solve for f(x) with x=1 I get ans =
    0.5460
I want to plug 0.5460 into a formula and get 1 back.
Chris, this is the Finv function you spoke of in my other question. 
 A: $$x = 1.003\sqrt{2}\left(\mathrm{erf}^{-1}\left(\frac{Y/1.42 + 0.4042}{0.5}\right) - 0.05\right) + 0.5501$$
A: Let's start with the "general form" you say you had: 
$$a\left(\frac12\mathrm{erf}\left(\frac{x-b}{c\sqrt{2}}+\frac12\right)\right)+d=y$$
and go over the inversion slo-o-owly...
$$a\left(\frac12\mathrm{erf}\left(\frac{x-b}{c\sqrt{2}}+\frac12\right)\right)=y-d$$
$$\frac12\mathrm{erf}\left(\frac{x-b}{c\sqrt{2}}+\frac12\right)=\frac{y-d}{a}$$
$$\mathrm{erf}\left(\frac{x-b}{c\sqrt{2}}+\frac12\right)=2\frac{y-d}{a}$$
Now, we can employ the inverse error function:
$$\frac{x-b}{c\sqrt{2}}+\frac12=\mathrm{erf}^{-1}\left(2\frac{y-d}{a}\right)$$
$$\frac{x-b}{c\sqrt{2}}=\mathrm{erf}^{-1}\left(2\frac{y-d}{a}\right)-\frac12$$
$$x-b=c\sqrt{2}\left(\mathrm{erf}^{-1}\left(2\frac{y-d}{a}\right)-\frac12\right)$$
$$x=b+c\sqrt{2}\left(\mathrm{erf}^{-1}\left(2\frac{y-d}{a}\right)-\frac12\right)$$
and that's the inverse you need.
Now, if it were
$$a\left(\frac12\mathrm{erf}\left(\frac{x-b}{c\sqrt{2}}\right)+\frac12\right)+d=y$$
instead (which I think is more likely, since you're starting from the normal distribution CDF), things are a bit different. What you should end up with is
$$x=b+c\sqrt{2}\,\mathrm{erf}^{-1}\left(2\frac{y-d}{a}-1\right)$$
which is probably the expression you actually need for those confidence intervals...
