There are an infinite number of solutions because you have one equation with 5 variables.
You could write (w+3)(2x-a) = bw^2 - w + 3c in terms of c, giving you this:
c = ( (w+3)(2x-a) - bw^2 + w) / 3
This would allow you to write a - b + c like this:
a - b + ( (w+3)(2x-a) - bw^2 + w) / 3
But since you have only 1 equation and 5 variables, you can only get rid of 1 of the variables. If you get rid of c, then you will (almost) always have a, b, w, and x left no matter what. If you get rid of a, then you will (almost) always have b, c, w, and x left no matter what.
There are certain cases where you might be able to get a - b + c in terms of w and x alone, but these are rare. In this case:
(w+3)(2x-a) = bw^2 - w + 3c
2xw - aw + 6x -3a = bw^2 - w + 3c
(-w-3)a + 2xw + 6x = (w^2)b + 3c - w
2xw + 6x - w = (w+3)a + (w^2)b + 3c
In order for this to allow a - b + c to be expressed in terms of just w and x, the coefficients in a - b + c (1, -1, and 1) would have to be proportional to the coefficients in (w+3)a + (w^2)b + 3c (w+3, w^2, and 3).
(1, -1, 1) can't be proportional to (w+3, w^2, 3) because the -1 and the (second) 1 in (1, -1, 1) are always opposite signs and the w^2 and 3 in (w+3, w^2, 3) are always the same sign.
So, without at least 2 additional equations (to knock 2 more variables), a - b + c can't be expressed in terms of just w and x.