d: \(\text{Δ}=\left(2m+3\right)^2-4\left(-2m-4\right)\)
\(=4m^2+12m+9+8m+16\)
\(=4m^2+20m+25\)
\(=\left(2m+5\right)^2>=0\)
=>Phương trình luôn có hai nghiệm
Ta có: \(\left|x_1-x_2\right|=5\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=25\)
\(\Leftrightarrow\left(2m+3\right)^2-4\left(-2m-4\right)=25\)
\(\Leftrightarrow4m^2+12m+9+8m+16=25\)
=>4m(m+5)=0
=>m=0 hoặc m=-5

