I can't solve this
For an integer n, the inequality x^2 + nx + 15 < -21 - 280
has no real solutions in x. Find the number of different possible values of n.
Simplify as
x^2 + nx + 316 < 0
Set as an equality
We will have real solutions whenever the discriminant is ≥ 0
So
n^2 - 4(1)(316) ≥ 0
n^2 - 1264 ≥ 0
n^2 ≥ 1264 take both roots
n ≥ sqrt (1264) → ≈ 35.6 or n ≤ -sqrt (1264) → ≈ -35.6
Thus....we have real solutions when n = ( inf , ≈ -35.6] U [ ≈ 35.6 , inf )
So the integers for n that make this have NO real solutions in x are the integers between these two values
n = [ -35 , 35 ] = 2(35) + 1 = 71 integers