Question 422669
You made a mistake in thinking that {{{(3y+2)(3y+2)}}} multiplies out to {{{9y^2+4}}}; however, it really multiplies out to {{{9y^2+12y+4}}}



{{{18y^2+24y+8=0}}} Start with the given equation.



Notice that the quadratic {{{18y^2+24y+8}}} is in the form of {{{Ay^2+By+C}}} where {{{A=18}}}, {{{B=24}}}, and {{{C=8}}}



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



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



{{{y = (-(24) +- sqrt( (24)^2-4(18)(8) ))/(2(18))}}} Plug in  {{{A=18}}}, {{{B=24}}}, and {{{C=8}}}



{{{y = (-24 +- sqrt( 576-4(18)(8) ))/(2(18))}}} Square {{{24}}} to get {{{576}}}. 



{{{y = (-24 +- sqrt( 576-576 ))/(2(18))}}} Multiply {{{4(18)(8)}}} to get {{{576}}}



{{{y = (-24 +- sqrt( 0 ))/(2(18))}}} Subtract {{{576}}} from {{{576}}} to get {{{0}}}



{{{y = (-24 +- sqrt( 0 ))/(36)}}} Multiply {{{2}}} and {{{18}}} to get {{{36}}}. 



{{{y = (-24 +- 0)/(36)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{y = (-24 + 0)/(36)}}} or {{{y = (-24 - 0)/(36)}}} Break up the expression. 



{{{y = (-24)/(36)}}} or {{{y =  (-24)/(36)}}} Combine like terms. 



{{{y = -2/3}}} or {{{y = -2/3}}} Simplify. 



So the solution is {{{y = -2/3}}}



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim