Question 375574
First, let's complete the square.



{{{3x^2-2x+7}}} Start with the given expression.



{{{3(x^2-(2/3)x+7/3)}}} Factor out the {{{x^2}}} coefficient {{{3}}}. This step is very important: the {{{x^2}}} coefficient <font size=4><b>must</b></font> be equal to 1.



Take half of the {{{x}}} coefficient {{{-2/3}}} to get {{{-1/3}}}. In other words, {{{(1/2)(-2/3)=-1/3}}}.



Now square {{{-1/3}}} to get {{{1/9}}}. In other words, {{{(-1/3)^2=(-1/3)(-1/3)=1/9}}}



{{{3(x^2-(2/3)x+highlight(1/9-1/9)+7/3)}}} Now add <font size=4><b>and</b></font> subtract {{{1/9}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{1/9-1/9=0}}}. So the expression is not changed.



{{{3((x^2-(2/3)x+1/9)-1/9+7/3)}}} Group the first three terms.



{{{3((x-1/3)^2-1/9+7/3)}}} Factor {{{x^2-(2/3)x+1/9}}} to get {{{(x-1/3)^2}}}.



{{{3((x-1/3)^2+20/9)}}} Combine like terms.



{{{3(x-1/3)^2+3(20/9)}}} Distribute.



{{{3(x-1/3)^2+20/3}}} Multiply.



So after completing the square, {{{3x^2-2x+7}}} transforms to {{{3(x-1/3)^2+20/3}}}. So {{{3x^2-2x+7=3(x-1/3)^2+20/3}}}.



So {{{3x^2-2x+7=0}}} is equivalent to {{{3(x-1/3)^2+20/3=0}}}.



Now let's solve for {{{(x-1/3)^2}}}



{{{3(x-1/3)^2+20/3=0}}} Start with the given equation.



{{{3(x-1/3)^2=-20/3}}} Subtract {{{20/3}}} from both sides.



{{{(x-1/3)^2=(-20/3)(1/3)}}} Multiply both sides by {{{1/3}}}.



{{{(x-1/3)^2=-20/9}}} Multiply



So the solution is {{{(x-1/3)^2=-20/9}}}



So you were close. You were just off by a sign. 



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Jim