The answer is YES. Let three nonzero real numbers are a,b,c.I'm going to prove by contradiction.Namely,we suppose that $ax^2+bx+c=0$ has only nonpositive solutions.Let these solutions are α≦0,β≦0.Then b=－a（α＋β）, c=aαβ.Since c≠0, α≠0,β≠0.So a,b,c are same sign.Let p,q,r which are obtained by rearrangement a,b,c are satisfied |p|≧|q|≧|r|＞0.Then $r^2-4pq<0$ which is absurd.Therefore the conclusion holds.