To pursue your idea of using a "difference of two squares", we can start by writing the quadratic equation in "vertex form" thus,
$$ ax^2 \ + \ bx \ + \ c \ \ = \ \ a·\left(x + \frac{b}{2a}\right)^2 \ + \ \left(\underbrace{c \ - \ \frac{b^2}{4a}}_{k} \right) $$ $$ = \ \ a·\left(x + \frac{b}{2a}\right)^2 \ - \ \left( \frac{b^2}{4a} \ - \ c \right) \ \ = \ \ 0 \ \ , $$
with the polynomial coefficients all being real numbers.
The equation will have non-real roots if the vertex is "above" the $ \ x-$axis $ \ ( k > 0 ) \ $ for $ \ a > 0 \ \ $ and "below" that axis $ \ ( k < 0 ) \ $ when $ \ a < 0 \ \ ; $ this is equivalent to the statement $ \ b^2 - 4ac < 0 \ $ in regard to the discriminant of the quadratic polynomial.
The difference of terms can then be factored as
$$ a \ · \ \left[ \ \left(x + \frac{b}{2a}\right)^2 \ - \ \left( \ \sqrt{ \underbrace{\frac{b^2 \ - \ 4ac}{4a^2}}_{< \ 0}} \ \right)^2 \ \right] $$ $$ = \ \ a \ · \left[ \ \left(x + \frac{b}{2a}\right) \ + \ i· \frac{\sqrt{4ac \ - \ b}} {2a} \ \right] · \left[ \ \left(x + \frac{b}{2a}\right) \ - \ i· \frac{\sqrt{4ac \ - \ b}} {2a} \ \right] \ \ = \ \ 0 \ \ , $$
from which it follows that the zeroes of the quadratic polynomial form the "complex conjugate pair" $ \ -\frac{b}{2a} \ \pm \ i· \frac{\sqrt{4ac \ - \ b}} {2a} \ \ . $