Rozwiązanie:
[tex]x^{2}+(m-4)x+m+2=0\\\Delta>0\\\Delta=m^{2}-8m+16-4(m+2)=m^{2}-8m+16-4m-8=m^{2}-12m+8\\m^{2}-12m+8>0\\\Delta_{m}=144-4*1*8=112\\m_{1}=\frac{12-4\sqrt{7} }{2} =6-2\sqrt{7}\\m_{2}=\frac{12+4\sqrt{7} }{2}=6+2\sqrt{7}\\m \in (-\infty, 6-2\sqrt{7}) \cup ( 6+2\sqrt{7},\infty) \\x_{1}+x_{2}<0\\4-m<0\\m>4\\x_{1}x_{2}>0\\m+2>0\\m>-2[/tex]
Ostatecznie otrzymamy:
[tex]m \in (6+2\sqrt{7} ,\infty)[/tex]
