Question 241759
{{{5y(y + 9) = 6(y + 9)}}} Start with the given equation.



{{{5y^2+45y = 6y + 54}}} Distribute



{{{5y^2+45y-6y-54=0}}} Get every term to the left side.



{{{5y^2+39y-54=0}}} Combine like terms.



Notice that the quadratic {{{5y^2+39y-54}}} is in the form of {{{Ay^2+By+C}}} where {{{A=5}}}, {{{B=39}}}, and {{{C=-54}}}



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



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



{{{y = (-(39) +- sqrt( (39)^2-4(5)(-54) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=39}}}, and {{{C=-54}}}



{{{y = (-39 +- sqrt( 1521-4(5)(-54) ))/(2(5))}}} Square {{{39}}} to get {{{1521}}}. 



{{{y = (-39 +- sqrt( 1521--1080 ))/(2(5))}}} Multiply {{{4(5)(-54)}}} to get {{{-1080}}}



{{{y = (-39 +- sqrt( 1521+1080 ))/(2(5))}}} Rewrite {{{sqrt(1521--1080)}}} as {{{sqrt(1521+1080)}}}



{{{y = (-39 +- sqrt( 2601 ))/(2(5))}}} Add {{{1521}}} to {{{1080}}} to get {{{2601}}}



{{{y = (-39 +- sqrt( 2601 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{y = (-39 +- 51)/(10)}}} Take the square root of {{{2601}}} to get {{{51}}}. 



{{{y = (-39 + 51)/(10)}}} or {{{y = (-39 - 51)/(10)}}} Break up the expression. 



{{{y = (12)/(10)}}} or {{{y =  (-90)/(10)}}} Combine like terms. 



{{{y = 6/5}}} or {{{y = -9}}} Simplify. 



So the solutions are {{{y = 6/5}}} or {{{y = -9}}}