Question 221482

{{{y^2-4y+4=9}}} Start with the given equation.



{{{y^2-4y+4-9=0}}} Subtract 9 from both sides.



{{{y^2-4y-5=0}}} Combine like terms.



Notice that the quadratic {{{y^2-4y-5}}} is in the form of {{{Ay^2+By+C}}} where {{{A=1}}}, {{{B=-4}}}, and {{{C=-5}}}



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



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



{{{y = (-(-4) +- sqrt( (-4)^2-4(1)(-5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-4}}}, and {{{C=-5}}}



{{{y = (4 +- sqrt( (-4)^2-4(1)(-5) ))/(2(1))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{y = (4 +- sqrt( 16-4(1)(-5) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{y = (4 +- sqrt( 16--20 ))/(2(1))}}} Multiply {{{4(1)(-5)}}} to get {{{-20}}}



{{{y = (4 +- sqrt( 16+20 ))/(2(1))}}} Rewrite {{{sqrt(16--20)}}} as {{{sqrt(16+20)}}}



{{{y = (4 +- sqrt( 36 ))/(2(1))}}} Add {{{16}}} to {{{20}}} to get {{{36}}}



{{{y = (4 +- sqrt( 36 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (4 +- 6)/(2)}}} Take the square root of {{{36}}} to get {{{6}}}. 



{{{y = (4 + 6)/(2)}}} or {{{y = (4 - 6)/(2)}}} Break up the expression. 



{{{y = (10)/(2)}}} or {{{y =  (-2)/(2)}}} Combine like terms. 



{{{y = 5}}} or {{{y = -1}}} Simplify. 



So the solutions are {{{y = 5}}} or {{{y = -1}}} 

  
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