\(4x^2-12x+9=\left(5-x\right)^2\Leftrightarrow4x^2-12x+9=25-10x+x^2\)

\(\Leftrightarrow4x^2-12x+9-25+10x-x^2+\Leftrightarrow3x^2-2x-16=0\)

\(\Delta'=\left(-1\right)^2-3.\left(-16\right)=1+48=49>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x_1=\dfrac{1+\sqrt{49}}{3}=\dfrac{1+7}{3}=\dfrac{8}{3}\) \(4x^2-12x+9=\left(5-x\right)^2\Leftrightarrow4x^2-12x+9=25-10x+x^2\)

\(x_2=\dfrac{1-\sqrt{49}}{3}=\dfrac{1-7}{3}=\dfrac{-6}{3}=-2\)

vậy \(x=\dfrac{8}{3};x=-2\)