Algebra

First, let's move all terms to one side and simplify. We have

\(4x^2 + nx + 104 < 0 \)

 

Let's set this up as an equation. We have

\(4x^2 + nx + 104 = 0\)

 

This will have real solutions when

\(n^2 - 4 (4) (104) ≥ 0 \\ n^2 ≥ 1664\)

\(n ≥ 40.79 \\ n ≤ -40.79\)

 

This is when there is a real value. Which means that when 

\( n < 40.79 \\ n > -40.79\)

there will be no real solutions. Thus, we have 

\( n = [ -40 , 40 ] \)

There are 81 values of n. 

 

Thus our answer is 81 !

 

Thanks! :)

\(x^2 + nx + 15 < -89 - 3x^2\)

so i first brought everything to the left side to get \(4x^2 + nx + 104 < 0\)

then i divided by 4 to get\(x^2 + \frac{n}{4}x + 26 < 0\)

for x to have no real solutions the discriminant must be negative

so \(\frac{n^2}{16}-4(26)\) must be negative. the less than sign is useless since smaller right hand values make c greater

the greatest value of n that works is 40

but we also can have negatives, least of which is -40

so -40 to 40 is \(\boxed{81}\) numbers