If [tex]r,s[/tex] are roots to the quadratic, then we can factor the quadratic and write
[tex](x-r)(x-s)=0=x^2-(r+s)x+rs[/tex]
so that [tex]-(r+s)=b[/tex] and [tex]rs=c[/tex]. It's not clear whether there's enough information given here, though... I'm guessing you're given that [tex]b<0[/tex] and [tex]c<0[/tex], in which case you'd have [tex]rs>0[/tex], so the answer would be no.
