I know the simple interest formula is, $I = prt$, but I don't know the interest to be able to use that formula.
Here's the variables. Please help me with the formula.
$$p = \$92,000.00$$
$$t = 72 \text{ months}$$
$$ \mathrm{Pmt} = \$1461.00$$
$$r = ?$$
This is a simple interest loan paid monthly and the rate is a fixed annual percentage.
I suppose that you need to use the classical formula $$A = P\frac{i(1 + i)^n}{(1 + i)^n - 1}$$ where $A$ is the monthly payment amount, $P$ the amount of the loan, $i$ the monthly interest and $n$ the total number of payments.
Everything is simple to compute using this formula except the interest (otherwise you would not ask the question) and, as far as I know, only numerical methods would be able to solve the problem.
Since $i$ is small compared to $1$, we can use Taylor expansion around $i=0$ and get as an approximation $$A=\frac{P}{n}+\frac{ (n+1) P}{2 n}i+\frac{ \left(n^2-1\right) P}{12 n}i^2+O\left(i^3\right)$$ which reduces to quadratic equation in $i$. Applied to your numbers $(A=1461$, $P=92000$, $n=72)$, the positive root is given by $$i=\frac{4 \sqrt{1175358765}-125925}{2980225}\approx 0.00376113$$ Using this number would lead to $A=1460.98$ which is not bad.
A rigorous calculation, based on Newton method, would lead to $i\approx 0.00376161$.
You can notice that, neglecting the second order term of the expansion, we would have obtained $$i=\frac{1649}{419750}\approx 0.00392853$$ which is too high but already quite accurate.
If you want to stay simple, you could use the simplest Pade approximant and write $$A=P \frac{\frac{i (n+2)}{3 n}+\frac{1}{n}}{\frac{1}{6} i (1-n)+1}$$ For your numbers, this would give $$i=\frac{9894}{2635579}\approx 0.00375401$$
