a) \(x^2+4x+3=0\)
\(\Leftrightarrow x^2+x+3x+3=0\)
\(\Leftrightarrow x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
b) Thay \(x=2\) vào (1), ta được:
\(2^2-6.2+m=0\)
\(\Leftrightarrow4-12+m=0\)
\(\Leftrightarrow-12+m=0\)
\(\Leftrightarrow m=12\)