Odpowiedź:
[tex]x^2+(2m-3)x+m^2-2m-3=0\\ a=1, b=(2m-3), c=(m^2-2m-3)\\ \Delta=(2m-3)^2-4\cdot1\cdot(m^2-2m-3)=\\=4m^2-12m+9-4m^2+8m+12=\\=-4m+21\\ \Delta>0\iff \ -4m+21>0\ \to \ m<\frac{21}4 \ \to \ m\in(-\infty,\frac{21}4)\\ x_1+x_2<x_1\cdot x_2\\ \iff \frac{-b}a<\frac ca\\ \frac{-(2m-3)}1<\frac{m^2-2m-3}1\\ -2m+3<m^2-2m-3\\ -m^2<-6\\ m^2>6\ \\ |m|>\sqrt6\\ m>\sqrt6 \ \vee \ m<-\sqrt6\ \to \ m\in(-\infty,-\sqrt6)\cup(\sqrt6,+\infty)[/tex]
[tex]\left \{ {{\Delta>0} \atop {x_1+x_2<x_1\cdot x_2}} \right. \ \to \ \left \{ {{m\in(-\infty,\frac{21}4)} \atop {m\in(-\infty,-\sqrt6)\cup(\sqrt6,+\infty)}} \right. \to \\ \to \boxed{m\in(-\infty,-\sqrt6)\cup(\sqrt6,\frac{21}4)}[/tex]