Question 476848


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



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



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)(6) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-5}}}, and {{{C=6}}}



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



{{{x = (5 +- sqrt( 25-4(1)(6) ))/(2(1))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25-24 ))/(2(1))}}} Multiply {{{4(1)(6)}}} to get {{{24}}}



{{{x = (5 +- sqrt( 1 ))/(2(1))}}} Subtract {{{24}}} from {{{25}}} to get {{{1}}}



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



{{{x = (5 +- 1)/(2)}}} Take the square root of {{{1}}} to get {{{1}}}. 



{{{x = (5 + 1)/(2)}}} or {{{x = (5 - 1)/(2)}}} Break up the expression. 



{{{x = (6)/(2)}}} or {{{x =  (4)/(2)}}} Combine like terms. 



{{{x = 3}}} or {{{x = 2}}} Simplify. 



So the solutions are {{{x = 3}}} or {{{x = 2}}}