z^2+2z + 1 +5>0
(z+1)^2> -5
Which has no real solutions.
take square root of each side
z+1 < isqrt5 but sqrt5 has two roots, so
-isqrt5<z+1<+isqrr5
subtract 1 from all sides, and you have it.
-1-isqrt5<z<-1+isqrt5
I am surprised your teacher gave you this, as I suspect it is a little over your experience level. If the 6 had been -6, it would be different.
Solve inequality and put solution set in interval notation.
z^2+2z+6>0
2 answers
Actually, (z-1)^2 >= 0, so it is > -5 for all x!