Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions? –5 –3 3 4

b = -5

Step-by-step explanation:

You needy to calculate the Discriminant = b² - 4ac

If D is zero there is one real solution,

If D less then zero there are no real solutions

Only if D is bigger then zero, there are two real solutions.

x² - 5x + 5 D > 0 so 2 real solutions

see attachment.

x² - 3x + 5 D < 0 so no real solutions

x² + 3x + 5 D < 0 so no real solutions

x² + 4x + 5 D < 0 so no real solutions

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right