Question 598370

{{{x^2-5x+5=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-5x+5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-5}}}, and {{{C=5}}}



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



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



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



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



{{{x = (5 +- sqrt( 25-4(1)(5) ))/(2(1))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{x = (5 +- sqrt( 5 ))/(2(1))}}} Subtract {{{20}}} from {{{25}}} to get {{{5}}}



{{{x = (5 +- sqrt( 5 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (5+sqrt(5))/(2)}}} or {{{x = (5-sqrt(5))/(2)}}} Break up the expression.  



So the solutions are {{{x = (5+sqrt(5))/(2)}}} or {{{x = (5-sqrt(5))/(2)}}} 



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