Very easy
\(15-2x-x^2=0\)
\(\Leftrightarrow16-\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow16-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)