Equation to estimate the error of propagation or delta method I'm calculating S
Where

*

*S =  (A - B)/A

*A = 778 ± 30 (value ± percent error)

*B = 388 ± 12 (value ± percent error)

I want to report the error in S as S ± error.
I have checked several online resources. I found many equations of how the error of propagation is derived but I couldn't find a specific equation that help me estimate the error in S based on the errors of A and B?
I'll highly appreciate any suggestion what is the equation I should use to estimate the error in S based on its two components A and B?
UPDATE
In this webpage about propagation of errors.
https://www.lhup.edu/~dsimanek/scenario/errorman/propagat.htm
In 3.5 Examples
Example no (2) below is similar to my case but not the same. First, my question is about 2 variables only A and B. In addition, there is subtraction in the nominator in my case not summation. Yet, they have not estimated the correlation between the three parameters G , H and Z.

A quantity Q is calculated from the law:
Q = (G+H)/Z,
and the data is:
G = 20 ± 0.5
H = 16 ± 0.5
z = 106 ± 1.0

 A: You should be able to get at this by using the delta method. I've made a post here: estimation of population ratio using delta method
about the delta method on a ratio which you can use the argument to give you exactly what you need. The final result is: 
$$1-\frac{B}{A}=\frac{A-B}{A} \sim N\left(1-\frac{\mu_B}{\mu_A}, \frac{\sigma^2}{n}\right)$$ 
Here $\sigma^2 = \frac{\sigma_B^2}{n\mu_B^2} - 2\frac{\sigma_{BA}}{\mu_B^2}+\frac{\sigma^2_A\mu_B^2}{n\mu_B^4}$. As I mentioned in the comment you'll either need to estimate the covariance $\sigma_{AB}$ (equivalently the correlation $\rho$) or assume the covariance is 0. 
From the data you provided in this post and assuming a $1.96$ multiplier to get 95% confidence intervals. I get the following for the parameter estimates: 
$\mu_A =778$, $\mu_B=388$, $\sigma^2_A=15.3061^2$,$\sigma^2_B=6.1224489^2$. Then $1-\mu_B/\mu_A=0.5076$ and $\sigma^2/n =0.0004376/n$. You can then again use $1.96$ as a multiplier on the standard deviation $\sigma$ to get a 95% confidence interval for this quantity. Here in my calculation for $\sigma^2$ I'm assuming the covariance is 0. 
