ОДЗ 
0} \atop {x^2+x-6>0}} \right.<=>\left \{ {{x^2+2x-8>0} \atop {x^2+x-6>0}} \right.<=>\left \{ {{(x-2)(x+4)>0} \atop {(x-2)(x+3)>0}} \right.\\\ \\\ (-\infty;-4)\cup(2;+\infty) " alt="\left \{ {{2x^2+4x-16>0} \atop {x^2+x-6>0}} \right.<=>\left \{ {{x^2+2x-8>0} \atop {x^2+x-6>0}} \right.<=>\left \{ {{(x-2)(x+4)>0} \atop {(x-2)(x+3)>0}} \right.\\\ \\\ (-\infty;-4)\cup(2;+\infty) " align="absmiddle" class="latex-formula">
![2x^2+4x-16=x^2+x-6\\\ 2x^2+4x-16-x^2-x+6=0\\\ x^2+3x-10=0\\\ D=9+40=49\\\ x_1=\frac{-3+7}{2}=2\\\ x_2=\frac{-3-7}{2}=-5\\\](https://tex.z-dn.net/?f=2x%5E2%2B4x-16%3Dx%5E2%2Bx-6%5C%5C%5C+2x%5E2%2B4x-16-x%5E2-x%2B6%3D0%5C%5C%5C+x%5E2%2B3x-10%3D0%5C%5C%5C+D%3D9%2B40%3D49%5C%5C%5C+x_1%3D%5Cfrac%7B-3%2B7%7D%7B2%7D%3D2%5C%5C%5C+x_2%3D%5Cfrac%7B-3-7%7D%7B2%7D%3D-5%5C%5C%5C+ "2x^2+4x-16=x^2+x-6\\\ 2x^2+4x-16-x^2-x+6=0\\\ x^2+3x-10=0\\\ D=9+40=49\\\ x_1=\frac{-3+7}{2}=2\\\ x_2=\frac{-3-7}{2}=-5\\\")
x=2 не удовлетворяет ОДЗ
Ответ: -5