\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow2^2\left(2x+7\right)^2-3^2\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)

\(\Leftrightarrow\left[\left(4x+14\right)-\left(3x+9\right)\right]\left[\left(4x+14\right)+\left(3x+9\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\7x=-23\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

Vậy \(x=-5\) hoặc \(x=\dfrac{-23}{7}\)