complete the square to solve the equation below x^2+10x+10=14?

x² + 10x + 10 = 14
x² + 10x = 4
x² + 2(5)x + 5² = 4 + 5²
(x + 5)² = 29
x + 5 = √29 or x + 5 = -√29
x = -5 + √29 or x = -5 - √29

x^2 + 10x + 10 = 14
x^2 + 10x = 14 -10
x^2 + 10x = 4
(•.• By applying (a+b)^2 formula)
x^2 + 2(5)x + 5^2 = 4 + 5^2
(x + 5)^2 = 29
x + 5 = ±√(29)
x+ 5 = +√(29) => x = +√(29) - 5
x + 5 = -√(29) => x = -√(29) -5

x^2 + 10x + 10 = 14
x^2 + 10x - 4 = 0
(x + 5)^2 - 29 = 0
x
= -5 +/- √29

x² + 10x = 4
x² + 10x + 25 = 29
[ x + 5 ]² = 29
x + 5 = ± √29
x = - 5 ± √29

x^2 + 10x = 4
(x + 5)^2 - 5^2 = 4
(x + 5)^2 = 25 + 4
(x + 5)^2 = 29
x + 5 = +/- sqrt(29)
x = - 5 +/- sqrt(29)

x² + 10x + 10 = 14

isolate constant terms
x² + 10x = 14-10
x² + 10x = 4

complete the square
 coefficient of the x term: 10
 divide it in half: 5
 square it: 5²
 use 5² to complete the square:
x² + 10x + 5² = 4 + 5²
(x+5)² = 29

x+5 = ±√29
x = -5±√29

x^2+10x+10=14
(x^2 +10x) = (x+5)^2 - 25 (subtracted 25 to compensate for the 25 needed to
complete the square)
(x+5)^2 -25 + 10 = 14
(x+5)^2 -15 = 14
(x+5)^2 = 14 + 15
(x+5)^2 = 29
take the square root of both sides
(x+5) = +/- sqrt(29)
subtract 5 from both sides
===== Answer
x = -5 +/- sqrt(29)

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