`ĐKXĐ:x\inR`
`\sqrt{4x^2-12x+9}=6`
`⇔\sqrt{(2x-3)^2}=6`
`⇔|2x-3|=6`
`⇔` \(\left[ \begin{array}{l}2x-3=6\\2x-3=-6\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=9\\2x=-3\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\frac{9}{2}\\x=\frac{-3}{2}\end{array} \right.\)
Vậy `x={9/2,-3/2}`