Answer:
A and D
Step-by-step explanation:
Given
x² - 10x - 5 = 9 ( add 5 to both sides )
x² - 10x = 14
To compete the square
add (half the coefficient of the x- term )² to both sides
x² + 2(- 5)x + 25 = 14 + 25
(x - 5)² = 39 ← take the square root of both sides
x - 5 = ± [tex]\sqrt{39}[/tex] ( add 5 to both sides )
x = 5 ± [tex]\sqrt{39}[/tex]
Hence
x = 5 - [tex]\sqrt{39}[/tex] → A
x = 5 + [tex]\sqrt{39}[/tex] → D
