a, Ta có:
\(\Delta=\left[-\left(m+5\right)\right]^2-4\left(2m+6\right)\\ =m^2+10m+25-8m-24\\ =m^2+2m+1\\ =\left(m+1\right)^2\ge0\)
Vậy pt luôn có 2 nghiệm x1,x2
b, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=m+5\\x_1x_2=2m+6\end{matrix}\right.\)
\(x^2_1+x^2_2=13\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=13\\ \Leftrightarrow\left(m+5\right)^2-2\left(2m+6\right)=13\\ \Leftrightarrow m^2+10m+25-4m-12-13=0\\ \Leftrightarrow m^2+6m=0\\ \Leftrightarrow m\left(m+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=0\\m=-6\end{matrix}\right.\)