Explaining the e solution for the elusiviness

And is that I want to find the Y for the case of the X be 5, if I put the 5 in the place of the X 

I would have 0/0 that is a real number but is elusiviness number, so I still don't know what is the Y when the X is 5, so I would say that the Y is 'a' so the 'a' is the value for Y, 

making the fucntion I would find that squared(x) - ax + (5a - 25) = 0

so when I came to the end I know the result for X that is 5 the I am looking for Y that I call 'a' finding the X that is 5, because I know the formula, so in the end applying the X=5 I find the 'a' equal to 10.

And you remember that 'a' is the Y that I can't find because for X = 5 I would have 0/0, so the 'a' is the Y since the X = 5.

Rute Bezerra de Menezes Gondim