X+1/x-3+12/x+3=24/x2-9

$\frac{x+1}{x-3}+\frac{12}{x+3}=\frac{24}{x^{2} -9}\\\\\frac{x^{2}+3x+x+3+12x-36 }{(x-3)(x+3)}=\frac{24}{x^{2}-9 }\\\\\frac{x^{2}+16x-33 }{x^{2}-9 }=\frac{24}{x^{2}-9 } \\\\\left \{ {{x^{2}+16x-33=24 } \atop {x\neq-3;x\neq 3 }} \right.\\\\x^{2} +16x-57=0\\\\D=16^{2}-4*(-57)=256+228=484=22^{2}\\\\x_{1}=\frac{-16+22}{2}=3\\\\x_{2}=\frac{-16-22}{2}=-\frac{38}{2}=-19$

Ответ : - 19