I would appreciate it if someone helps me with solving the following equation. Suppose $\lambda,l \in R^+$, $p\in[0,1]$, and $s\in N_{0}$. How can we find an $x\in [0,l]$, which satisfies the following equation:
$$\sum_{z=0}^{s} \frac{(\lambda(l-x))^z}{z!} e^{-\lambda(l-x)}=p$$
or
$$F(s,\lambda(l-x))=p,$$ where $F(.,\lambda(l-x))$ is Poisson CDF with mean $\lambda(l-x)$.
On
Even if $x$ was not in the exponent, this would be a polynomial equation of degree $s$, there is no analytic solutions for $s \ge 5$. This way, it is nonlinear even for $s=1$.
However, because the LHS is continuous and differentiable in $x$, you can use a variety of numerical methods, like bisection or Newton's method, to numerically approximate the root if it exists.
Considering $$\sum_{z=0}^{s} \frac{(\lambda(l-x))^z}{z!} e^{-\lambda(l-x)}=p$$ I would suggest to change variable $\lambda(l-x)=t$ which makes $$\sum_{z=0}^{s}\frac{e^{-t} t^z}{z!}=p$$ which simplifies to $$\frac{\Gamma (s+1,t)}{s!}=p$$ where appears the incomplete Gamma function.
So, given $s$ and $p$, this is the equation to be solved. If $p\leq 1$, you will find a solution but you would more than likely use a numerical method as @gt6989b already answered.
Edit
Playing with the equation using $p=\frac 12$, it is amazing to find that the solution is almost $$t \approx s+\frac 23$$ For example, using $s=20$, this gives $t=\frac{62}{3}\approx 20.6667$ while the "exact" solution is $\approx 20.6676$.