=>(x-2)(x+2)+3(x-2)=0
=>(x-2)(x+5)=0
=>x=2 hoặc x=-5
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
theo đề ta có\(x^2-4+3\left(x-2\right)=0\)
\(x^2-4+3x-6=0\)
\(\Leftrightarrow x^2+3x=6+4=10\)
\(x\left(x+3\right)=10\)
\(\left\{{}\begin{matrix}x=10\\x+3=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\x=7\end{matrix}\right.\)
vậy x \(\in\left\{10,7\right\}\)
