Different method altogether, though I stand by the idea that Display name's method is the desired solution.
I call this the over/under method. For instance: x^2 + 3x + 2 = 0. Add to it (both sides) to get the square using one more "x"... more clearly stated, get x^2 + 4x + 4 = x + 2 ("Over"). Solving, the left side is (x + 2)^2 and divide each side by (x + 2) leaving x + 2 = 1, then solve to get x = –1.
Second step is the other direction, going down an "x": x^2 + 2x + 1 = –x – 1 with the left side being (x + 1)^2 ("Under"). Divide each side by (x + 1) to get x + 1 = –1 and solve to get x = –2.
So x = –1, –2. No discriminants, no factoring of the nasty, ugly kind. I.e.: the hard kind.
Over/under method. Nothing complicated, just add an x, subtract an x, complete each square, divide to get the squared term equals 1 or –1, solve each.
(For a ≠ 1, as in this problem, divide through by "a" first for the easy solution (c'mon, fractions aren't hard to work with), or just do the slightly harder completing of the square, if you like "harder.")