\(x^2-7x+12=0\)
=>(x-3)(x-4)=0
=>x=3 hoặc x=4
Theo đề, ta có: \(\left\{{}\begin{matrix}9-3a+b=0\\16-4a+b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3a+b=-9\\-4a+b=-16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=12\end{matrix}\right.\)