Me with this
plz explain

options for all are:
two real solutions
one real solution
no real solution

When you have a bunch like this, I find it easiest to graph them.

The numbers of real solutions are ...
  2
  0
  2
  1

I know you know how to use Desmos to graph these. You can also write a little spreadsheet program to compute the discriminant b^2 -4ac, or do it the hard way.
  1^2 -4(-3)(12) = 145 . . . positive, so 2 solutions
  (-6)^2 -4(2)(5) = -4 . . .. negative, so 0 solutions
  7^2 -4(1)(-11) = 93 . . . .. positive, so 2 solutions
  (-8)^2 -4(-1)(-16) = 0 . .. zero, so 1 solution

Given that: y = cos(2pi/5)x -1

The general form of cos function is y = A cos(Bx+C) +D

So amplitude = A=1

Period = 2pi/B=2pi/(2pi/5)= 5

Phase shift = -C/B= 0

Vertical shift = D=-1

So the equation of the midline is y=-1

Sot he correct option is C.