$\sqrt{4-6x-x^2}-x=4$
ĐKXĐ: $-4 \le x \le -3+\sqrt{13}$
$⇔ \sqrt{-x^2-6x+4}=x+4$
$⇔-x^2-6x+4 = x^2+8x+16$
$⇔ -2x^2-14x-12=0$
$⇔ -2(x^2+7x+6)=0$
$⇔ x^2+7x+6=0$
$⇔ (x+1)(x+6)=0$
\(⇔ \left[ \begin{array}{l}x=-1 \ (tm)\\x=-6\ (ktm)\end{array} \right.\)
Vậy $S=\{-1\}$