How do you solve $(x-2/3)^2 - 8= 1/3$ ?

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Answer 1 · 2017-11-28T00:40:32

$x_1=(2-5sqrt3)/3$ and $x_2=(2+5sqrt3)/3$

$(x-2/3)^2-8=1/3$

$(x-2/3)^2-25/3=0$

$(x-2/3)^2-75/9=0$

$(x-2/3)^2-((5sqrt3)/3)^2=0$

$(x-2/3+(5sqrt3)/3)*(x-2/3-(5sqrt3)/3)=0$

Hence $x_1=(2-5sqrt3)/3$ and $x_2=(2+5sqrt3)/3$

How do you solve (x-2/3)^2 - 8= 1/3 (x-2/3)^2 - 8= 1/3 ?