Ta có: \(\left(3x+5\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(3x+5\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=x-3\\3x+5=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-x=-5-3\\3x+x=3-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-8\\4x=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy x = 4 hoặc x =\(-\dfrac{1}{2}\).