This is the excersice:

$$x^2 - 2mx - 2m - 1 = 0$$

I've done this:

1.- $$x^2 - (2mx -2m) - 1 = 0$$
2.- $$x^2 - (2m(x-1)) - 1 = 0$$

I think I need to transform: 

$$(2m(x-1))$$ to something of the form $$mx$$

And then to have something like: 

$$x^2 - mx - 1 = 0$$

So after I can apply the quadratic formula, but not sure how to proceed. 

**UPDATE 1**:

After David answer, i have this:

$$x^2 - 2mx - (2m + 1)(-1) = 0$$

So: a -> 1, b -> -2m, and c -> (2m + 1)(-1). 

If it is correct, is right to assume that the answer marked as correct from "juantheron" is more practical?

I see the solution on the marked answer, but I want to understand every other possibilities.