Question 339009


{{{x^2-6x+13=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-6x+13}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-6}}}, and {{{C=13}}}



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)(13) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-6}}}, and {{{C=13}}}



{{{x = (6 +- sqrt( (-6)^2-4(1)(13) ))/(2(1))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{x = (6 +- sqrt( 36-4(1)(13) ))/(2(1))}}} Square {{{-6}}} to get {{{36}}}. 



{{{x = (6 +- sqrt( 36-52 ))/(2(1))}}} Multiply {{{4(1)(13)}}} to get {{{52}}}



{{{x = (6 +- sqrt( -16 ))/(2(1))}}} Subtract {{{52}}} from {{{36}}} to get {{{-16}}}



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



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



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



{{{x = (6)/(2) + (4*i)/(2)}}} or {{{x =  (6)/(2) - (4*i)/(2)}}} Break up the fraction for each case. 



{{{x = 3+2*i}}} or {{{x =  3-2*i}}} Reduce. 



{{{x = 3+2*i}}} or {{{x = 3-2*i}}} Simplify. 



So the solutions are {{{x = 3+2*i}}} or {{{x = 3-2*i}}} 

  


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