Решите неравенство (x+4)^2>16-x^2

$(x+4)^2\ \textgreater \ 16-x^2 \\ \\ x^2+ 8x+16 - 16 + x^2\ \textgreater \ 0 \\ \\ 2x^2+ 8x \ \textgreater \ 0 \\ \\ x(2x+ 8) \ \textgreater \ 0$

$\left \{ {{x \ \textgreater \ 0} \atop {2x+ 8 \ \textgreater \ 0}} \right. \ \ \ \left \{ {{x \ \textgreater \ 0} \atop {x \ \textgreater \ -4}} \right. \ \ \Rightarrow \ x\ \textgreater \ 0$

или

$\left \{ {{x \ \textless \ 0} \atop {2x+ 8 \ \textless \ 0}} \right. \left \{ {{x \ \textless \ 0} \atop {x \ \textless \ -4}} \right. \Rightarrow \ \ x \ \textless \ -4}$

Ответе: $x \ \textless \ -4; \ \ x \ \textgreater \ 0$