Question 554326
You would use the quadratic formula.



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



Notice that the quadratic {{{x^2-2x+2}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-2}}}, 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 = (-(-2) +- sqrt( (-2)^2-4(1)(2) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-2}}}, and {{{C=2}}}



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



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



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



{{{x = (2 +- sqrt( -4 ))/(2(1))}}} Subtract {{{8}}} from {{{4}}} to get {{{-4}}}



{{{x = (2 +- sqrt( -4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



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



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



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



{{{x = 1+i}}} or {{{x =  1-i}}} Reduce. 



So the solutions are {{{x = 1+i}}} or {{{x =  1-i}}}



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim