\(\Delta'=\left(m+1\right)^2-2m-10=m^2-9\)
Phương trình có hai nghiệm khi \(\Delta'\ge0\Leftrightarrow m^2-9\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge3\\m\le-3\end{matrix}\right.\)
Theo định lí Vi-ét \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=2m+10\end{matrix}\right.\)
Khi đó \(A=14x_1x_2+x_1^2+x_2^2=12x_1x_2+\left(x_1+x_2\right)^2\)
\(=24m+120+4m^2+8m+4\)
\(=4m^2+32m+124=4\left(m+4\right)^2+60\ge60\)
\(MinA=60\Leftrightarrow m=-4\left(tm\right)\)
Vậy \(m=-4\)