Question 150982


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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-5}}}, and {{{c=2}}}



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)(2) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-5}}}, and {{{c=2}}}



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



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



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



{{{x = (5 +- sqrt( 17 ))/(2(1))}}} Subtract {{{8}}} from {{{25}}} to get {{{17}}}



{{{x = (5 +- sqrt( 17 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (5+sqrt(17))/(2)}}} or {{{x = (5-sqrt(17))/(2)}}} Break up the expression.  



So our answers are {{{x = (5+sqrt(17))/(2)}}} or {{{x = (5-sqrt(17))/(2)}}} 



which approximate to {{{x=4.562}}} or {{{x=0.438}}}