Determine which functions have two real number zeros by calculating the discriminant, b2 – 4ac. check all that apply. a. f(x) = x2 + 6x + 8 b. g(x) = x2 + 4x + 8 c. h(x) = x2 – 12x + 32 d. k(x) = x2 + 4x – 1 e. p(x) = 5x2 + 5x + 4 f. t(x) = x2 – 2x – 15

f(x) = x^2 + 6x + 8= b^2 - 4ac= (6)^2 - 4(1)(8)= 36 - 4(8)= 36 - 32= 4
g(x) = x2 + 4x + 8= b^2 - 4ac= (4)^2 - 4(1)(8)= 16 - 4(8)= 16 - 32= -16
h(x) = x2 – 12x + 32= b^2 - 4ac= (-12)^2 - 4(1)(32)= 144 - 4(32)= 144 - 128= 16
k(x) = x2 + 4x – 1 
= (4)^2 - 4(1)(-1)= 16 - 4(-1)= 16 + 4= 20
p(x) = 5x2 + 5x + 4= b^2 - 4ac= (5)^2 - 4(5)(4)= 25 - 4(20)= 25 - 80= -55
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than do 23.50 times 18 and then divide your answer by 100 than you will get x.