Question

how many solutions does b^2 - 4ac < 0 have

Answer 1

I need a bit of clarification — do you mean:

1) "If b^2 − 4ac < 0, how many solutions does the quadratic equation ax^2 + bx + c = 0 have?" — or
2) "How many triples (a,b,c) satisfy the inequality b^2 − 4ac < 0?"

Answers for both:

1) For the quadratic ax^2 + bx + c = 0 with real coefficients and a ≠ 0: the discriminant is D = b^2 − 4ac. If D < 0 there are no real roots; there are two distinct complex conjugate roots given by
x = (−b ± sqrt(D)) / (2a),
and since D is negative the square root is imaginary, so you get two distinct complex (nonreal) solutions.

2) As an inequality in the real variables a,b,c, the set { (a,b,c) ∈ R^3 : b^2 − 4ac < 0 } is an open region of R^3, so there are infinitely many such triples (uncountably many). (If you restrict to integers or other domains, the answer depends on that restriction.)

If you meant something else, tell me the context and I’ll answer accordingly.