Theo đề, ta có:
\(\left\{{}\begin{matrix}a+c+2=-4\\-\dfrac{4-4ac}{4a}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-6-c\\4ac-4=24a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-6-c\\4c\left(-6-c\right)-4-24\left(-6-c\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-c-6\\-24c+4c^2-4+144+24c=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-c-6\\4c^2+140=0\end{matrix}\right.\)(vô lý)