Question 526292


{{{x^2-6x+5=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-6x+5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-6}}}, 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 = (-(-6) +- sqrt( (-6)^2-4(1)(5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-6}}}, and {{{C=5}}}



{{{x = (6 +- sqrt( (-6)^2-4(1)(5) ))/(2(1))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{x = (6 +- sqrt( 36-4(1)(5) ))/(2(1))}}} Square {{{-6}}} to get {{{36}}}. 



{{{x = (6 +- sqrt( 36-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{x = (6 +- sqrt( 16 ))/(2(1))}}} Subtract {{{20}}} from {{{36}}} to get {{{16}}}



{{{x = (6 +- sqrt( 16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (6 +- 4)/(2)}}} Take the square root of {{{16}}} to get {{{4}}}. 



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



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



{{{x = 5}}} or {{{x = 1}}} Simplify. 



So the solutions are {{{x = 5}}} or {{{x = 1}}}