Question 595870

{{{x^2-10x+29=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-10x+29}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-10}}}, and {{{C=29}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-10) +- sqrt( (-10)^2-4(1)(29) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-10}}}, and {{{C=29}}}



{{{x = (10 +- sqrt( (-10)^2-4(1)(29) ))/(2(1))}}} Negate {{{-10}}} to get {{{10}}}. 



{{{x = (10 +- sqrt( 100-4(1)(29) ))/(2(1))}}} Square {{{-10}}} to get {{{100}}}. 



{{{x = (10 +- sqrt( 100-116 ))/(2(1))}}} Multiply {{{4(1)(29)}}} to get {{{116}}}



{{{x = (10 +- sqrt( -16 ))/(2(1))}}} Subtract {{{116}}} from {{{100}}} to get {{{-16}}}



{{{x = (10 +- sqrt( -16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (10 +- 4*i)/(2)}}} Take the square root of {{{-16}}} to get {{{4*i}}}. 



{{{x = (10 + 4*i)/(2)}}} or {{{x = (10 - 4*i)/(2)}}} Break up the expression. 



{{{x = (10)/(2) + (4*i)/(2)}}} or {{{x =  (10)/(2) - (4*i)/(2)}}} Break up the fraction for each case. 



{{{x = 5+2*i}}} or {{{x =  5-2*i}}} Reduce. 



So the solutions are {{{x = 5+2*i}}} or {{{x = 5-2*i}}}