Từ đề bài ta có \(a\ne0\) và:
\(\left\{{}\begin{matrix}-\frac{b}{2a}=1\\\frac{4ac-b^2}{4a}=4\\a-b+c=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\c=1-3a\\4ac-b^2=16a\end{matrix}\right.\)
\(\Rightarrow4a\left(1-3a\right)-4a^2=16a\)
\(\Rightarrow-16a=12\Rightarrow a=-\frac{3}{4}\) ; \(b=\frac{3}{2}\) ; \(c=\frac{13}{4}\)