Question 384096
Yes. From the quadratic formula, we have

{{{x = (-b +- sqrt(b^2 - 4ac))/2a}}}

Suppose the discriminant was negative. Then, this value x becomes

{{{(-b +- di)/2a = (-b/2a) +- (d/2a)i}}} where d is real. We can assign arbitrary values u and v to obtain

x = {{{u +- vi}}}.