Question 538704


{{{ sqrt(6y+4)+6=y }}} Start with the given equation.


{{{ sqrt(6y+4)=y-6 }}} Subtract 6 from both sides.


{{{ (sqrt(6y+4))^2=(y-6)^2 }}} Square both sides


{{{ 6y+4=y^2-12y+36 }}} Square and FOIL


{{{ 0=y^2-12y+36-6y-4 }}} Get everything to one side


{{{ 0=y^2-18y+32 }}} Combine like terms.


{{{ y^2-18y+32=0 }}} Flip the equation



Notice that the quadratic {{{y^2-18y+32}}} is in the form of {{{Ay^2+By+C}}} where {{{A=1}}}, {{{B=-18}}}, and {{{C=32}}}



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



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



{{{y = (-(-18) +- sqrt( (-18)^2-4(1)(32) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-18}}}, and {{{C=32}}}



{{{y = (18 +- sqrt( (-18)^2-4(1)(32) ))/(2(1))}}} Negate {{{-18}}} to get {{{18}}}. 



{{{y = (18 +- sqrt( 324-4(1)(32) ))/(2(1))}}} Square {{{-18}}} to get {{{324}}}. 



{{{y = (18 +- sqrt( 324-128 ))/(2(1))}}} Multiply {{{4(1)(32)}}} to get {{{128}}}



{{{y = (18 +- sqrt( 196 ))/(2(1))}}} Subtract {{{128}}} from {{{324}}} to get {{{196}}}



{{{y = (18 +- sqrt( 196 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (18 +- 14)/(2)}}} Take the square root of {{{196}}} to get {{{14}}}. 



{{{y = (18 + 14)/(2)}}} or {{{y = (18 - 14)/(2)}}} Break up the expression. 



{{{y = (32)/(2)}}} or {{{y =  (4)/(2)}}} Combine like terms. 



{{{y = 16}}} or {{{y = 2}}} Simplify. 



So the possible solutions are {{{y = 16}}} or {{{y = 2}}} 



However, when doing the check (which I'll let you do), you'll find that only {{{y=16}}} works.



So {{{y=16}}} is the only solution. 



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