What are the zeros of the function f (x) = (x+2) ^2-25

F (x) = (x + 2) ² - 25

f (x) = (x + 2) (x + 2) - 25

f (x) = x (x + 2) + 2 (x + 2) - 25

f (x) = x (x) + x (2) + 2 (x) + 2 (2) - 25

f (x) = x² + 2x + 2x + 4 - 25

f (x) = x² + 4x + 4 - 25

f (x) = x² + 4x - 21

0 = x² + 4x - 21

x = - (4) ± √ ((4) ² - 4 (1) (-21))

2 (1)

x = - 4 ± √ (16 + 84)

2

x = - 4 ± √ (100)

2

x = - 4 ± 10

2

x = - 2 ± 5

x = - 2 + 5 U x = - 2 - 5

x = 3 U x = - 7

The solution set is {3, - 7}.