Bài 3:
a: Ta có: \(4x\left(x-7\right)-4x^2=56\)
\(\Leftrightarrow4x^2-28x-4x^2=56\)
\(\Leftrightarrow-28x=56\)
hay x=-2
b: Ta có: \(12x\left(3x-2\right)+6x-4=0\)
\(\Leftrightarrow\left(3x-2\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{1}{6}\end{matrix}\right.\)