Suppose we are given $a,b,z \in \mathbb{R}$ with $a<b$ and $0\le z\le 1$. Can we approximate the roots of the following function?
$$f(x)=\text{erf}(b-x)-\text{erf}(a-x)-z$$
where $\text{erf}$ is the error function $\text{erf}(y)=\frac{2}{\sqrt{\pi}}\int_{0}^{y}e^{-t^2}dt$.
Background: I am working on solving a constrained integral of a Gaussian distribution with $\sigma^2=\frac{1}{2}$ over a given range [a,b], where $\frac{z}{2}$ is the value of the equality constraint.
What I tried so far: If $z=0$, I get $$\displaylines{\text{erf}(a-x)=\text{erf}(b-x)\\\ a-x=b-x\\\ x=\pm \infty}$$ but I didn't make any progress if $z\neq 0$.
Assuming that $a$ and $b$ are "close" to eachother $$f(x)=\text{erf}(b-x)-\text{erf}(a-x)-z$$ The first derivative cancels at $$x_*=\frac {a+b}2$$ and the second derivative test shows that it is a maximum.
Expanding as a Taylor series $$z=-2 \text{erf}\left(\frac{a-b}{2}\right)+\frac{2 (a-b) e^{-\frac{1}{4} (a-b)^2}}{\sqrt{\pi }}\left(x-\frac{a+b}{2}\right)^2+$$ $$\frac{\left((a-b)^2-6\right) (a-b) e^{-\frac{1}{4} (a-b)^2}}{6 \sqrt{\pi }}\left(x-\frac{a+b}{2}\right)^4+O\left(\left(x-\frac{a+b}{2}\right)^6\right)$$ which is just a quadratic equation in $\left(x-\frac{a+b}{2}\right)^2$