Answer:
x is 2 and -8
Step-by-step explanation:
[tex] {x}^{2} + 6x - 16 = 0[/tex]
general equation
[tex] {ax}^{2} + (sum)x + product = 0[/tex]
sum is 6, product is -16
for completing squares,
first divide the sum by 2:
[tex] = \frac{6}{2} = 3[/tex]
add the square of the result on (x² + 6x) and subtract it from the product:
[tex] ( {x}^{2} + 6x + {3}^{2} ) - 16 - {3}^{2} = 0 \\ {(x + 3)}^{2} - 25 = 0 \\ {(x + 3)}^{2} = 25[/tex]
take square root:
[tex] \sqrt{ {(x + 3)}^{2} } = \sqrt{25 } \\ x + 3 = ±5 \\ x = ±5 - 3[/tex]
x is either: 5-3 or -5-3
[tex]x = 2 \: \: and \: - 8[/tex]
