X(2х-1)-(6+х)(х-6)+(х+10)(10-х) = x(2х-1)-(х+6)(х-6)+(10+х)(10-х ) =
= 2х²-х-х²+36+100-х²=136-х
х=-101
136-(-101)=136+101=237
A=-1 b=-3 c=18
D=b2-4ac
D=9-4×-1×18
D=81
x1,2=
=-b+-√D
--------------
2a
=3-9
-----------
-2
=3
3+9
---------
-2
=-6
![x^{2} -3x-10<0](https://tex.z-dn.net/?f=x%5E%7B2%7D+-3x-10%3C0)
![x^{2} -3x-10=0](https://tex.z-dn.net/?f=x%5E%7B2%7D+-3x-10%3D0)
![x_{1} =-2](https://tex.z-dn.net/?f=x_%7B1%7D+%3D-2)
![x_{2} =5](https://tex.z-dn.net/?f=x_%7B2%7D+%3D5)
![(x+2)(x-5)<0](https://tex.z-dn.net/?f=%28x%2B2%29%28x-5%29%3C0)
Точки
и ![-5](https://tex.z-dn.net/?f=-5)
∈![(-5;2)](https://tex.z-dn.net/?f=%28-5%3B2%29)
-----------------------------------------
![x^{2} -25>0](https://tex.z-dn.net/?f=x%5E%7B2%7D+-25%3E0)
![(x-5)(x+5)>0](https://tex.z-dn.net/?f=%28x-5%29%28x%2B5%29%3E0)
Точки
и ![5](https://tex.z-dn.net/?f=5)
x∈(-∞;-5) ∪ (5;+∞)
вроде так