How do I find the solution set of 2≤x2+x≤122≤x2+x≤12?

Answer 1

2 < x^2 + x < 12

0 < x^2 + x - 2 < 10

x^2 + x - 2 = 0

( x + 2)(x - 1 ) = 0

x = -2 or x = 1

The min occurs at -b/(2a) = -1/(2*1) = -1/2.

and y = (-1/2)^2 + (-1/2) - 2 = 1/4 - 1/2 - 2 = -2.25

So this quadratic function parabola is positive when x<-2 and x>1