Hello from MrBillDoesMath!
Answer: 20 (the first choice)
Discussion:
Recall the solutions of the quadratic equation: a x^2 + bx + c = 0 are
( -b +\- sqrt ( b^2 - 4ac) ) /2a.
If there are no real solutions, then b^2 - 4ac must be < 0. In our case a = 1, b = 4, and c = c! For no real solutions this becomes
(4)^2 - 4 (1) c < 0 or
16 -4c < 0 or
16 < 4c (and dividing both sides by 4)
c > 4.
Of the choices provided only the first one, 20, is > 4.
Thank you,
MrB
