Ta có: \(a-b+c=4-\left(m^2+2m-15\right)+\left(m+1\right)^2-20=0\)
\(\Rightarrow\) Phương trình đã cho luôn có 2 nghiệm: \(\left[{}\begin{matrix}x=-1\\x=\dfrac{20-\left(m+1\right)^2}{4}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x_1=-1\\x_2=\dfrac{20-\left(m+1\right)^2}{4}\end{matrix}\right.\)
\(\Rightarrow1+\dfrac{20-\left(m+1\right)^2}{4}+2019=0\)
\(\Rightarrow\left(m+1\right)^2=8100\)
\(\Rightarrow\left[{}\begin{matrix}m=-91\\m=89\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x_1=\dfrac{20-\left(m+1\right)^2}{4}\\x_2=-1\end{matrix}\right.\)
\(\Rightarrow\left[\dfrac{20-\left(m+1\right)^2}{4}\right]^2+2020=0\)
\(\Leftrightarrow\left[\dfrac{20-\left(m+1\right)^2}{4}\right]^2=-2020\) (vô nghiệm)
Vậy \(m=\left\{-91;89\right\}\)
