$(3x+1)^2-(3x+1)(4x+4)+4(x+1)^2=25(x+3)^2\\↔(3x+1)^2-4(x+1)(3x+1)^2+4(x+1)^2=25(x+3)^2\\↔[(3x+1)-2(x+1)]=25(x+3)^2\\↔(3x+1-2x-2)=[5(x+3)]^2\\↔(x-1)^2=(5x+15)^2\\↔(x-1)^2-(5x+15)^2=0\\↔[(x-1)-(5x+15)][(x-1)+(5x+15)]=0\\↔(x-1-5x-15)(x-1+5x+15)=0\\↔(-4x-16)(6x+14)=0\\↔\left[\begin{array}{1}-4x-16=0\\6x+14=0\end{array}\right.\\↔\left[\begin{array}{1}4x=-16\\6x=-14\end{array}\right.\\↔\left[\begin{array}{1}x=-4\\x=-\dfrac{7}{3}\end{array}\right.$
Vậy $x∈\left\{-4;-\dfrac{7}{3}\right\}$