Referring to the same formula from this question (Compound Interest Formula adding annual contributions).
$$ F=P(1+i)^n+A\frac{(1+i)^n-1}{i}=\left(P+\frac{A}{i}\right)(1+i)^n-\frac{A}{i} $$
where
P = Principle value
i = interest rate, annually
A = Annual amount contribution
n = duration in years
F = Future value
If all of the variables above is known, except i. How do we get the formula of i? I'm having trouble in getting the formula of i.
$$ i=? $$
For this kind of problem, as soon as $n>4$, there is no analytical solution since the problem would be to find the roots of a polynomial of degree $n$ on $i$.
So, either numerical methods or approximations.
For approximations, since $i$ is known to be small, use for example a series expansion, which would give $$F-(P+A n)=\sum_{k=1}^\infty \Bigg(A \binom{n}{k+1}+P \binom{n}{k}\Bigg)\,i^k$$
Using only the first term,it will provide a starting point for any numerical method such as Newton $$i_0=\frac 2n\,\frac{F-(P+A n) }{(n-1)A+2P }$$
Using, for illustration : $F=2500$, $P=1000$, $A=100$ and $n=10$, this would give $i_0=\frac 1{29}=0.0344828$ while the solution is $i=0.0305279$.
Newton iterates would be $$\left( \begin{array}{cc} k & i_k \\ 0 & 0.0344828 \\ 1 & 0.0305879 \\ 2 & 0.0305280 \\ 3 & 0.0305279 \\ \end{array} \right)$$
which is a quite fast convergence.
If you do not want to use numerical methods, let $$B=F-(P+A n) \qquad \text{and} \qquad a_k=A \binom{n}{k+1}+P \binom{n}{k}$$ which make $$B=\sum_{k=1}^\infty a_k\, i^k$$ Truncate to some order and use power series reversion.
This would give $$i \sim \frac 1{a_1}B-\frac {a_2}{a_1^3}B^2+\frac{2a_2^2 -a_1a_3}{a_1^5} B^3+\cdots$$ and you can use as many terms as you want.
Using the same numbers and the above very truncated series would give $$i \sim \frac{629221}{20511149}=0.030677$$ Adding the next term would give $i=0.0304949$.