Question 627057


{{{y^2-9y-10=0}}} Start with the given equation.



Notice that the quadratic {{{y^2-9y-10}}} is in the form of {{{Ay^2+By+C}}} where {{{A=1}}}, {{{B=-9}}}, and {{{C=-10}}}



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



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



{{{y = (-(-9) +- sqrt( (-9)^2-4(1)(-10) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-9}}}, and {{{C=-10}}}



{{{y = (9 +- sqrt( (-9)^2-4(1)(-10) ))/(2(1))}}} Negate {{{-9}}} to get {{{9}}}. 



{{{y = (9 +- sqrt( 81-4(1)(-10) ))/(2(1))}}} Square {{{-9}}} to get {{{81}}}. 



{{{y = (9 +- sqrt( 81--40 ))/(2(1))}}} Multiply {{{4(1)(-10)}}} to get {{{-40}}}



{{{y = (9 +- sqrt( 81+40 ))/(2(1))}}} Rewrite {{{sqrt(81--40)}}} as {{{sqrt(81+40)}}}



{{{y = (9 +- sqrt( 121 ))/(2(1))}}} Add {{{81}}} to {{{40}}} to get {{{121}}}



{{{y = (9 +- sqrt( 121 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (9 +- 11)/(2)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{y = (9 + 11)/(2)}}} or {{{y = (9 - 11)/(2)}}} Break up the expression. 



{{{y = (20)/(2)}}} or {{{y =  (-2)/(2)}}} Combine like terms. 



{{{y = 10}}} or {{{y = -1}}} Simplify. 



So the solutions are {{{y = 10}}} or {{{y = -1}}} 



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