As you know, there is no explicit formula for the interest rate and numerical methods are required.
Starting from @Quanto's answer $$P =L\frac{r(1+r)^{n}}{(1+r)^{n}-1}$$ define $k=\frac PL$ and, rearranging, consider that you look for the zero of function
$$f(r)=(k-r)(1+r)^n-k$$
$$f'(r)=-(r+1)^{n-1} ((n+1) r+1-kn)$$ Starting from a reasonable guess $r_0$, Newton method will generates the interates according to
$$r_{m+1}=r_m-\frac {f(r_m)}{f'(r_m)}$$
Now, in orser to save iterations, we need a "good" estimate of $r_0$. Since $r \ll 1$, expand $f(r)$ as a Taylor series to get
$$f(r)=r (k n-1)+\frac{1}{2} r^2 \left(k n^2-k n-2 n\right)+O\left(r^3\right)$$ Ignoring the higher order terms as well as the trivial $r=0$, we have
$$r_0=\frac{2 (k n-1)}{n (2-k(n-1))}$$
For illustartion purposes, let us use $n=180$, $L=100000$ and $A=700$. This will give as estimate
$$x_0=\frac{26}{6723}\approx 0.00386732$$ while the exact solution would be $0.00266252$. Now, using Newton method, the generated iterates will be
$$\left(
\begin{array}{cc}
m & r_m \\
0 & 0.00386732 \\
1 & 0.00304223 \\
2 & 0.00271802 \\
3 & 0.00266399 \\
4 & 0.00266252
\end{array}
\right)$$
For sure, we can improve the quality of the estimate. For example, using more terms in the Taylor series and using series reversion, we could have
$$r_0=t-\frac{n-1}6 t^2+\frac{2n^2-n-1}{36 }t^3-\frac{22 n^3+3 n^2-18 n-7}{1080}t^4 \qquad \text{where} \qquad t=\frac{2 (k n-1)}{n+1}$$ For the worked example, this would give
$$r_0=\frac{1204948442464117}{452791316671875000}\approx0.00266116$$
For sure, we could continue adding terms and get an estimate closer and closer to the solution.
