Question 264730
{{{sqrt(3y-2)=y-2}}} Start with the given equation.



{{{3y-2=(y-2)^2}}} Square both sides to eliminate the square root.



{{{3y-2=y^2-4y+4}}} FOIL



{{{0=y^2-4y+4-3y+2}}} Get all terms to one side.



{{{0=y^2-7y+6}}} Combine like terms.



Notice that the quadratic {{{y^2-7y+6}}} is in the form of {{{Ay^2+By+C}}} where {{{A=1}}}, {{{B=-7}}}, and {{{C=6}}}



Let's use the quadratic formula to solve for "y":



{{{y = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{y = (-(-7) +- sqrt( (-7)^2-4(1)(6) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-7}}}, and {{{C=6}}}



{{{y = (7 +- sqrt( (-7)^2-4(1)(6) ))/(2(1))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{y = (7 +- sqrt( 49-4(1)(6) ))/(2(1))}}} Square {{{-7}}} to get {{{49}}}. 



{{{y = (7 +- sqrt( 49-24 ))/(2(1))}}} Multiply {{{4(1)(6)}}} to get {{{24}}}



{{{y = (7 +- sqrt( 25 ))/(2(1))}}} Subtract {{{24}}} from {{{49}}} to get {{{25}}}



{{{y = (7 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (7 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{y = (7 + 5)/(2)}}} or {{{y = (7 - 5)/(2)}}} Break up the expression. 



{{{y = (12)/(2)}}} or {{{y =  (2)/(2)}}} Combine like terms. 



{{{y = 6}}} or {{{y = 1}}} Simplify. 



So the <i>possible</i> solutions are {{{y = 6}}} or {{{y = 1}}} 



However, we need to check them. To do that, simply plug each possible solution into {{{sqrt(3y-2)=y-2}}}. If you get a true equation, then it will verify that particular possible solution. I'll let you do the check. Hint: One of the "possible" solutions is not a solution at all.