Answer:
we conclude that:
![x^2+6x+9>2x^2+14\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1](https://tex.z-dn.net/?f=x%5E2%2B6x%2B9%3E2x%5E2%2B14%5Cquad%20%3A%5Cquad%20%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3A1%3Cx%3C5%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%5Cleft%281%2C%5C%3A5%5Cright%29%5Cend%7Bbmatrix%7D)
Hence, (1, 5) is the solution in interval notation.
Please also check the attached graph.
Step-by-step explanation:
Given the inequality expression
![\left(x+3\right)^2>\:2\left(x^2+7\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2B3%5Cright%29%5E2%3E%5C%3A2%5Cleft%28x%5E2%2B7%5Cright%29)
as
(x + 3)² = x² + 6x + 9
2(x² + 7) = 2x² + 14
so
![\:x^2+6x+9\:>\:2x^2+14](https://tex.z-dn.net/?f=%5C%3Ax%5E2%2B6x%2B9%5C%3A%3E%5C%3A2x%5E2%2B14)
rewriting in the standard form
![-x^2+6x-5>0](https://tex.z-dn.net/?f=-x%5E2%2B6x-5%3E0)
Factor -x² + 6x - 5: - (x - 1) (x - 5)
![-\left(x-1\right)\left(x-5\right)>0](https://tex.z-dn.net/?f=-%5Cleft%28x-1%5Cright%29%5Cleft%28x-5%5Cright%29%3E0)
Multiply both sides by -1 (reverse the inequality)
![\left(-\left(x-1\right)\left(x-5\right)\right)\left(-1\right)](https://tex.z-dn.net/?f=%5Cleft%28-%5Cleft%28x-1%5Cright%29%5Cleft%28x-5%5Cright%29%5Cright%29%5Cleft%28-1%5Cright%29%3C0%5Ccdot%20%5Cleft%28-1%5Cright%29)
Simplify
![\left(x-1\right)\left(x-5\right)](https://tex.z-dn.net/?f=%5Cleft%28x-1%5Cright%29%5Cleft%28x-5%5Cright%29%3C0)
so
![1](https://tex.z-dn.net/?f=1%3Cx%3C5)
Therefore, we conclude that:
![x^2+6x+9>2x^2+14\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1](https://tex.z-dn.net/?f=x%5E2%2B6x%2B9%3E2x%5E2%2B14%5Cquad%20%3A%5Cquad%20%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3A1%3Cx%3C5%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%5Cleft%281%2C%5C%3A5%5Cright%29%5Cend%7Bbmatrix%7D)
Hence, (1, 5) is the solution in interval notation.
Please also check the attached graph.