Question 629850
{{{3x^2-6x+1=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2-6x+1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-6}}}, and {{{C=1}}}



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)(1) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-6}}}, and {{{C=1}}}



{{{x = (6 +- sqrt( (-6)^2-4(3)(1) ))/(2(3))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{x = (6 +- sqrt( 36-4(3)(1) ))/(2(3))}}} Square {{{-6}}} to get {{{36}}}. 



{{{x = (6 +- sqrt( 36-12 ))/(2(3))}}} Multiply {{{4(3)(1)}}} to get {{{12}}}



{{{x = (6 +- sqrt( 24 ))/(2(3))}}} Subtract {{{12}}} from {{{36}}} to get {{{24}}}



{{{x = (6 +- sqrt( 24 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



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



{{{x = (3+sqrt(6))/(3)}}} or {{{x = (3-sqrt(6))/(3)}}} Reduce.  



So the solutions are {{{x = (3+sqrt(6))/(3)}}} or {{{x = (3-sqrt(6))/(3)}}}


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