2(x-5)^2=3 solve for x......would it be 25/3 to solve for x........please help?

Question

x^2+4x+4=7 would I do this problem the same way……….

Completing the square x^2-6x-3=0 I know I have find the root first……..

Completing the square 2x^2+10+11=0 please help me…….I am struggling……

Answer 1

2(x-5)^2 = 3
divide by 2: (x - 5)^2 = 3/2
square root: (x - 5) = +/- root(6)/2
add 5: x = 5 +/- root(6)/2

Answer 2

First divide both sides by 2, so you have (x-5)^2 = 3/2. Then take the square root of both sides, giving you (x-5) = sqrt(3/2). Add five to both sides, giving you x = sqrt(3/2) + 5.

Answer 3

2(x - 5)^2 = 3
(x - 5)^2 = (3/2)
x - 5 = sqrt(3/2)
x - 5 = (sqrt(6))/2
x = 5 ± (1/2)sqrt(6)

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x^2 + 4x + 4 = 7
(x + 2)^2 = 7
x + 2 = ±sqrt(7)
x = -2 ± sqrt(7)

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x^2 - 6x - 3 = 0
x^2 - 6x = 3
find half of 6, square it, add to both sides
x^2 - 6x + 9 = 12
(x - 3)^2 = 12
x - 3 = sqrt(12)
x - 3 = sqrt(4 * 3)
x - 3 = 2sqrt(3)
x = 3 ± 2sqrt(3)

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2x^2 + 10x + 11 = 0
x^2 + 5x + (11/2) = 0
x^2 + 5x = (-11/2)
x^2 + 5x + (25/4) = (3/4)
(x + (5/2))^2 = (3/4)
x + (5/2) = sqrt(3/4)
x + (5/2) = (1/2)sqrt(3)
x = (-5/2) ± (1/2)sqrt(3)

Answer 4

First, divide both sides by 2 to get (x-5)^2 = 3/2
The square root both sides to get x-5 = + or - sqrt(3/2)

Then add 5 to both sides to get either 5 + sqrt(3/2) or 5 - sqrt(3/2).

Answer 5

2(x-5)^2=3
2(x^2-10x+25)=3
2x^2-20x+50=3
2x^2-20x+47=0
x=(20 +- sqrt((-20)^2-4(2)(47))/2(2)
x=(20 +- sqrt(400-376))/4
x=(20 +- sqrt(24))/4
x=6.2247 or 3.7753

Answer 6

(x-5)2=3/2
x-5=sqrt(3/2)
x=5-sqrt(3/2)

Answer 7

duhhh x is 8

Answer 8

(x-5)^2 = 1.5
x - 5 = +/- sqrt (1.5)
x = 5 +/- sqrt (1.5) = approximately 6.22 or 3.78