$\displaystyle \begin{array}{{>{\displaystyle}l}} \vartriangle '=1+m^{2} +4=m^{2} +5 >0\\ \Rightarrow PT\ luôn\ có\ 2\ n_{o} \ phân\ biệt\\ Theo\ Viet:\ x_{1} +x_{2} =2\ \ \ \ x_{1} x_{2} =-m^{2} -4\\ Có\ x_{1} -x_{2} =10\\ \Rightarrow ( x_{1} -x_{2})^{2} =100\\ \Leftrightarrow x_{1}^{2} +x_{2}^{2} -2x_{1} x_{2} =100\\ \Leftrightarrow ( x_{1} +x_{2})^{2} -4x_{1} x_{2} =100\\ \Leftrightarrow 2^{2} +4\left( m^{2} +4\right) =100\\ \Leftrightarrow m=\pm 2\sqrt{5} \end{array}$
