Question 715029

(Edited to permit reading symbolism more easily)
{{{(3y-1)^(2) =5}}}

Taking square roots to start, there will be two kinds; one positive and one negative:

{{{3y-1=+sqrt(5)}}}
{{{3y=1+sqrt(5)}}}
{{{y=(1+sqrt(5))/3}}}

OR


{{{3y-1=-sqrt(5)}}}
{{{3y=1-sqrt(5)}}}
{{{y=(1-sqrt(5))/3}}}


Final Answer: {{{highlight(y=(1-sqrt(5))/3)}}}  or  {{{highlight(y=(1+sqrt(5))/3)}}}