Question 323274

{{{x^2-5x+3=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-5x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-5}}}, and {{{C=3}}}



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)(3) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-5}}}, and {{{C=3}}}



{{{x = (5 +- sqrt( (-5)^2-4(1)(3) ))/(2(1))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{x = (5 +- sqrt( 25-4(1)(3) ))/(2(1))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25-12 ))/(2(1))}}} Multiply {{{4(1)(3)}}} to get {{{12}}}



{{{x = (5 +- sqrt( 13 ))/(2(1))}}} Subtract {{{12}}} from {{{25}}} to get {{{13}}}



{{{x = (5 +- sqrt( 13 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (5+sqrt(13))/(2)}}} or {{{x = (5-sqrt(13))/(2)}}} Break up the expression.  



So the solutions are {{{x = (5+sqrt(13))/(2)}}} or {{{x = (5-sqrt(13))/(2)}}} 



which approximate to {{{x=4.303}}} or {{{x=0.697}}}