Question 266157


{{{y^2-10y+25=16}}} Start with the given equation.



{{{y^2-10y+25-16=0}}} Subtract 16 from both sides.



{{{y^2-10y+9=0}}} Combine like terms.



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



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



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



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



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



{{{y = (10 +- sqrt( 100-4(1)(9) ))/(2(1))}}} Square {{{-10}}} to get {{{100}}}. 



{{{y = (10 +- sqrt( 100-36 ))/(2(1))}}} Multiply {{{4(1)(9)}}} to get {{{36}}}



{{{y = (10 +- sqrt( 64 ))/(2(1))}}} Subtract {{{36}}} from {{{100}}} to get {{{64}}}



{{{y = (10 +- sqrt( 64 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (10 +- 8)/(2)}}} Take the square root of {{{64}}} to get {{{8}}}. 



{{{y = (10 + 8)/(2)}}} or {{{y = (10 - 8)/(2)}}} Break up the expression. 



{{{y = (18)/(2)}}} or {{{y =  (2)/(2)}}} Combine like terms. 



{{{y = 9}}} or {{{y = 1}}} Simplify. 



So the solutions are {{{y = 9}}} or {{{y = 1}}}