jmc

Let there be $x$ rows with $y$ band members per row in Formation $A$. Then we have $xy=105$. From the information about Formation $B$, we have $(x+6)(y-2)=105$. Multiplying and substituting $xy=105$, we get $xy-2x+6y-12=105-2x+6y-12=105\iff -2x+6y=12\iff x=3y-6$. We now substitute $3y-6$ for $x$ in $xy=105$ and use the quadratic formula to solve the resulting quadratic equation, $3y^2-6y-105=0$. The positive solution, $y=7$, gives $x=15$. Thus, there are $\boxed{15}$ rows in Formation $A$.