Question 242287


{{{2x^2-5x+1=0}}} Start with the given equation.



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



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



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



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



{{{x = (5 +- sqrt( 25-8 ))/(2(2))}}} Multiply {{{4(2)(1)}}} to get {{{8}}}



{{{x = (5 +- sqrt( 17 ))/(2(2))}}} Subtract {{{8}}} from {{{25}}} to get {{{17}}}



{{{x = (5 +- sqrt( 17 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (5+sqrt(17))/(4)}}} or {{{x = (5-sqrt(17))/(4)}}} Break up the expression.  



So the solutions are {{{x = (5+sqrt(17))/(4)}}} or {{{x = (5-sqrt(17))/(4)}}} 



which approximate to {{{x=2.281}}} or {{{x=0.219}}}