solve this radical?

Question

9(x-3)^2-5=0
solve in square roots

Answer 1

9(x - 3)^2 - 5 = 0

9(x - 3)^2 = 5

(x - 3)^2 = 5/9

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

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

Answer 2

this is one way:
9(x-3)^2 - 5 = 0
9(x-3)^2 = 5
(add 5 to both sides)
(x-3)^2 = 5/9
(divide both sides by 9)
x-3 = + or - the square root of 5/9
(because (x-3)^2 is = to 5/9, the square root of 5/9 therefore must = (x-3). Due to the fact that x could be positive or negative we must say that the square root of 5/9 can be + or -)
x = (+ or - the suare root of 5/9) +3
(add 3 to both sides)

hope u can make sense of that

Answer 3

first 9(x - 3)^2 - 5 = 0
bring five over
9(x - 3)^2 = 5
divide by nine
(x - 3)^2 = 5/9
then square root it
x - 3 = +/- sqrt(5)/3
and add 3
x = 3 +/- sqrt(5)/3

Answer 4

9(x-3)^2=5

(x-3)^2=5/9

x-3= plus minus sqroot(5/9)
x= plus minus sqroot(5/9)+3