Do pt có nghiệm (1;-2) nên ta có:
\(\left\{{}\begin{matrix}1-2b=-2\\b+2a=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2b=3\\a=\dfrac{-3-b}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{2}\\a=-\dfrac{9}{4}\end{matrix}\right.\)