Question

(2n+7)<(n+3)^2
A.P.
it's important .....m​

Answer 1

Answer:

hey mate your answer is...

Step-by-step explanation:

(2n+7)<(n+3)^2

=2n+7<n^2+6n+9

=2n+7-n^2-6n-9<0

=-4n-2-n^2<0

=n^2-4n-2=0

n=-2+√2

n=-2-2√2.

hope it's helps.....

Answer 2

$(2n+7)<(n+3)^2\\2n+7<n^2+6n+9\\n^2+4n+2>0\\n^2+4n+4-2>0\\(n+2)^2>2\\n+2>\sqrt2 \vee n+2<-\sqrt2\\n>-2+\sqrt2 \vee n<-2-\sqrt2\\n\in(-\infty,-2-\sqrt2)\cup(-2+\sqrt2,\infty)$