Question 548728


{{{x^2-3x=-3}}} Start with the given equation.



{{{x^2-3x+3=0}}} Get every term to the left side.



Notice that the quadratic {{{x^2-3x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=3}}}



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



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



{{{x = (-(-3) +- sqrt( (-3)^2-4(1)(3) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=3}}}



{{{x = (3 +- sqrt( (-3)^2-4(1)(3) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(1)(3) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9-12 ))/(2(1))}}} Multiply {{{4(1)(3)}}} to get {{{12}}}



{{{x = (3 +- sqrt( -3 ))/(2(1))}}} Subtract {{{12}}} from {{{9}}} to get {{{-3}}}



{{{x = (3 +- sqrt( -3 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (3 +- i*sqrt(3))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (3+i*sqrt(3))/(2)}}} or {{{x = (3-i*sqrt(3))/(2)}}} Break up the expression.  



So the solutions are {{{x = (3+i*sqrt(3))/(2)}}} or {{{x = (3-i*sqrt(3))/(2)}}} 



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