![(2x^2+11x+6)(2x^2+11x+13)<8;\\ y=2x^2+11x+6;\\ y(y+7)<8;\\ y^2+7y-8<0;\\ (y+8)(y-1)<0;\\ -8<2x^2+11x+6<1<y (2x^2+11x+6)(2x^2+11x+13)<8;\\ y=2x^2+11x+6;\\ y(y+7)<8;\\ y^2+7y-8<0;\\ (y+8)(y-1)<0;\\ -8<2x^2+11x+6<1<y](https://tex.z-dn.net/?f=%282x%5E2%2B11x%2B6%29%282x%5E2%2B11x%2B13%29%3C8%3B%5C%5C+y%3D2x%5E2%2B11x%2B6%3B%5C%5C+y%28y%2B7%29%3C8%3B%5C%5C+y%5E2%2B7y-8%3C0%3B%5C%5C+%28y%2B8%29%28y-1%29%3C0%3B%5C%5C+-8%3C2x%5E2%2B11x%2B6%3C1%3Cy)
решим первое из двойного неравенства
0;\\ 2x^2+11x+14>0;\\ D=11^2-4*2*14=9=3^2;\\ x_1=\frac{-11-3}{2*2}=-3.5;\\ x_2=\frac{-11+3}{2*2}=-2;\\ 2(x+3.5)(x+2)>0" alt="-8<2x^2+11x+6;\\ 2x^2+11x+6+8>0;\\ 2x^2+11x+14>0;\\ D=11^2-4*2*14=9=3^2;\\ x_1=\frac{-11-3}{2*2}=-3.5;\\ x_2=\frac{-11+3}{2*2}=-2;\\ 2(x+3.5)(x+2)>0" align="absmiddle" class="latex-formula">
x є ![(-\infty; -3.5) \cup (-2; +\infty) (-\infty; -3.5) \cup (-2; +\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%3B+-3.5%29+%5Ccup+%28-2%3B+%2B%5Cinfty%29)
решим второе из двойного неравенства
![2x^2+11x+6<1;\\ 2x^2+11x+6-1<0;\\ 2x^2+11x+5<0;\\ D=11^2-4*2*5=81=9^2;\\ x_1=\frac{-11-9}{2*2}=-5;\\ x_2=\frac{-11+9}{2*2}=-0.5 2x^2+11x+6<1;\\ 2x^2+11x+6-1<0;\\ 2x^2+11x+5<0;\\ D=11^2-4*2*5=81=9^2;\\ x_1=\frac{-11-9}{2*2}=-5;\\ x_2=\frac{-11+9}{2*2}=-0.5](https://tex.z-dn.net/?f=2x%5E2%2B11x%2B6%3C1%3B%5C%5C+2x%5E2%2B11x%2B6-1%3C0%3B%5C%5C+2x%5E2%2B11x%2B5%3C0%3B%5C%5C+D%3D11%5E2-4%2A2%2A5%3D81%3D9%5E2%3B%5C%5C+x_1%3D%5Cfrac%7B-11-9%7D%7B2%2A2%7D%3D-5%3B%5C%5C+x_2%3D%5Cfrac%7B-11%2B9%7D%7B2%2A2%7D%3D-0.5)
x э (-5; -0.5)
обьединяя общие решения получаем
х є ![(-5;-3.5) \cup (-2;-0.5) (-5;-3.5) \cup (-2;-0.5)](https://tex.z-dn.net/?f=%28-5%3B-3.5%29+%5Ccup+%28-2%3B-0.5%29+)
ответ: ![(-5;-3.5) \cup (-2;-0.5) (-5;-3.5) \cup (-2;-0.5)](https://tex.z-dn.net/?f=%28-5%3B-3.5%29+%5Ccup+%28-2%3B-0.5%29+)