Need a to be nonzero, as someone else hinted at.

You can use the fundamental theorem of algebra, which says that every polynomial with complex coefficients has at least one complex root. Probably a bit overkill since it's a quadratic. That said, here's a slightly roundabout way of thinking about it:

Say you have a quadratic polynomial P(x). You know it has some complex root, call it x_0. In other words, P(x_0) = 0. You can prove that this is equivalent to the statement that (x - x_0) divides P(x). In other words,

P(x) = Q(x)(x - x_0),

where Q(x) is another polynomial with complex coefficients. But you know P(x) has degree 2, and the degree of the product of two polynomials is the sum of their individual degrees. Thus given that (x - x_0) has degree 1, you know Q(x) must be a degree 1 polynomial also. Factoring out whatever the leading coefficient Q(x) is will give you the desired result.