Find Interest RATE of Mortgage When Got Loan & Payment I can seem to find an answer to this question. It sounds like it should be simple but i can't find antying online and only found one answer on stack that seem overly complicated. I have a loan amount, term and total payment and i need to simply solve for the interst rate. 
I am trying to simple get the interest rate of a mortgage loan (amortized) when loan amount, term and PI payment is know. Now i want to use that information to solve for the interest rate. 
loan amount , term, payment = rate???

Can someone please help me solve this equation in a step by step manner. The only solution i found online was some exhausting formulas with no explanation and I couldn't make out the variables or workflow because there was not step by step calculation.
Thank you.
 A: The rate on your loan should be somewhere on your statement.
The basic formula
$B = P \frac {(1+y)^N + 1}{y(1+y)^N}$
$B$ is the balance on your loan.
$P$ is your payment.
$N$ is the number of payments remaining.
$y$ is your periodic rate.  So, if you are making monthly payments. $y = \frac {r}{12}$ with r as your annualized rate.
Rather than trying to find a formula for $y,$ I suggest you numerical methods to find a solution.
A: Here is how to derive the equation among the load amount $L$, the payment $P$, the term $N$ and the rate $r$. Since the amount for each payment is the same, equalize the n-th and (n+1)-th payments 
$$rA_n+X_n = r(A_n - X_n) + X_{n+1}\tag{1}$$
Where $A_n$ is the principal outstanding at time $n$ and $X_n$ is the principal amortized at time $n$. Simplify (1) to get $X_{n+1} = (1+r)X_n$, The total amount amortized throughout the term is $L$, i.e.
$$L = X_1 [1+(1+r)+(1+r)^2+\>…\>..+(1+r)^{N-1}]$$
Sum up the geometric series and express
$$ X_1=\frac{rL}{(1+r)^{N}-1}$$
Then, substitute $X_1$ into the payment $P = (1+r)L+X_1$ to obtain the sought-after equation,
$$P =L\frac{r(1+r)^{N}}{(1+r)^{N}-1} $$
Note that the formula derived is applicable for annual payments. The monthly version is similarly derived, only with $r$ placed by $\frac r{12}$.
A: 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.
