\(\Delta=m^2-4.\left(-1\right)=m^2+4>0\)
\(\hept{\begin{cases}x_1=\frac{-m-\sqrt{m^2+4}}{2}\\x_2=\frac{-m+\sqrt{m^2+4}}{2}\end{cases}}\)
Để x1<2
\(\Rightarrow m+\sqrt{m^2+4}>-4\)
Có\(\sqrt{m^2+4}\ge\sqrt{4}=2\)
\(\Rightarrow m+2>-4\)
\(\Leftrightarrow m>-6\)
Vậy m>-6 để....