Question 152737


{{{3x^2-6x+3=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=-6}}}, 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 = (-(-6) +- sqrt( (-6)^2-4(3)(3) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-6}}}, and {{{c=3}}}



{{{x = (6 +- sqrt( (-6)^2-4(3)(3) ))/(2(3))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{x = (6 +- sqrt( 36-4(3)(3) ))/(2(3))}}} Square {{{-6}}} to get {{{36}}}. 



{{{x = (6 +- sqrt( 36-36 ))/(2(3))}}} Multiply {{{4(3)(3)}}} to get {{{36}}}



{{{x = (6 +- sqrt( 0 ))/(2(3))}}} Subtract {{{36}}} from {{{36}}} to get {{{0}}}



{{{x = (6 +- sqrt( 0 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (6 +- 0)/(6)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (6 + 0)/(6)}}} or {{{x = (6 - 0)/(6)}}} Break up the expression. 



{{{x = (6)/(6)}}} or {{{x =  (6)/(6)}}} Combine like terms. 



{{{x = 1}}} or {{{x = 1}}} Simplify. 



So the only solution is {{{x = 1}}}