$a,\ (x+4)-(x+1)(x-1)=16$
$⇔x+4-x^2+1=16$
$⇔x^2-x+11=0$
$⇔(x^2-2.\frac{1}{2}x+\frac{1}{4})+\frac{43}{4}=0$
$⇔(x-\frac{1}{2})^2+\frac{43}{4}=0$
Vì $(x-\frac{1}{2})^2≥0∀x⇒(x-\frac{1}{2})^2+\frac{43}{4}>0∀x$
Vậy phương trình đã cho vô nghiệm
$b,\ (2x-1)^2+(x+3)^2-5(x+7)(x-7)=6$
$⇔4x^2-4x+1+x^2+6x+9-5x^2+245=6$
$⇔2x=-249$
$⇔x=\frac{-249}{2}$
Vậy $S=\{\frac{-249}{2}\}$.