\(\Delta=\left(m+2\right)^2-4\left(m-1\right)=m^2+2m+8=\left(m+1\right)^2+7>0\)
Phương trình luôn có 2 nghiệm phân biệt
Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m-2\\x_1x_2=m-1\end{matrix}\right.\)
\(x_1^2+x_2^2-13=x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-3x_1x_2-13=0\)
\(\Leftrightarrow\left(-m-2\right)^2-3\left(m-1\right)-13=0\)
\(\Leftrightarrow m^2+m-6=0\Rightarrow\left[{}\begin{matrix}m=2\\m=-3\end{matrix}\right.\)