How to calculate monthly interest payment from loan? My professor put out a piece of text that says this is how you calculate the monthly interest payment on a loan:

but I can't arrive at the monthly interest payment.
When I type in:
(.00417*2000)/((1-(1+.00417) - 24))

I get -0.347439...
What am I doing wrong?
 A: The actual formula is
$$P = \frac{iA}{1-(1+i)^{-n}} \tag{1}\label{eq1}$$
As shown above, plus as the J. W. Tanner and saulspatz question comments indicated, the "$-n$" at the end of it was poorly formatted and was actually meant to be a power, not subtracting $n$ from the amount. Using $i = .00417$, $n = 24$ and $A = 2000$, \eqref{eq1} gives a monthly payment amount of
$$P = \frac{0.00417 \times 2000}{1 - (1.00417)^{-24}} = \$87.746\ldots \tag{2}\label{eq2}$$
Although it appears the official answer truncates down to the nearest cent, note that $i = .05/12 = .0041\bar{6}$, with using this more accurate value giving $P = \$87.742\ldots$, so just normal rounding is needed & was likely used. The total payments would be $\$87.74 \times 24 = \$2,105.76$, so the total interest paid over $2$ years would be $\$2,105.76 - \$2,000.00 = \$105.76$.
The Web page Amortization Calculation Formula has the formula of
$$A = P \frac{r(1 + r)^n}{(1 + r)^n - 1} \tag{3}\label{eq3}$$
Here, the $r$ is your $i$, plus the $A$ and $P$ symbols are switched around. When you divide the numerator and denominator by $(1 + r)^n$, you then get your \eqref{eq1} equivalent formula.
A: Another intuition to find the monthly payment for a loan is to take the present value of all monthly payments and equate it with the loan amount itself, or in formula: 
\begin{align}  A = P  \cdot a_{\overline{n}|} \end{align}
in which $A$ is the loan amount, $P$ is the monthly payment,  and 
$ a_{\overline{n}|} $ is just the present value of the annuity formula, which is 
\begin{align} a_{\overline{n}|} = \frac{1-\frac{1}{(1+i)^{n}}}{i} \end{align}
This boils down to 
\begin{align} 
P  = \frac{A}{a_{\overline{n}|}}  
   = \frac{A}{\frac{1-\frac{1}{(1+i)^{n}}}{i}} 
   = \frac{iA}{1-\frac{1}{(1+i)^{n}}}
   = \frac{iA}{1-(1+i)^{-n}}
\end{align} 
which is the same exact formula as the one your professor gave you.
