I need to explain what 2 reasons why has no real zeros. im very confused anything helps :)

Answer in explanation

Step-by-step explanation:

The graph of f does not cross the x-axis. We know this because for any value x, we always get that x^2+4 is positive. x^2 is positive or zero ( 0 or greater than 0) for any x but adding 4 to it makes It 4 or greater than 4.

The discriminant is negative.

That is, b^2-4ac is negative.

Upon comparing ax^2+bx+c to x^2+4 we see that a=1, b=0, and c=4..

Plugging those into b^2-4ac gives (0)^2-4(1)(4)=0-16=-16.

Negative discriminate implies two nonreal zeros so it will not cross the x-axis.

(Positive discriminant implies 2 real and zero discriminate imples 1 real.)

answer: -8.2562234e+13

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3