find the exact values when :

Question

x^2+2x >= 5

Answer 1

x^2+2x-5=>0

Well, the issue is what are the factors on the left.

Using the quadratic equation

x=(-2+- sqrt (4+4*5))/2=-1+-sqrt6

(x+1-sqrt6)(x+1+sqrt6)>=0
Well, for the left side to be > 0, both factors have to be positive or negative.

Take x=-100000. That works. So negative x have to be considered.
Case I
x+1-sqrt 6<0 And x+1+sqrt6<0
x<-1+sqrt6 AND x<=-1-sqrt6
Since it has to be AND
then x<=-1-sqrt6
Now, when are both positive:
Case II
x+1-sqrt6>=0 AND x+1-sqrt6>=0
x>-1+sqrt6 AND x>-1+sqrt6
Since that is AND, then
x>-1+sqrt6

So finally, the exact values are

-1+sqrt6<x OR
x<-1-sqrt6

Check my logic.