document.write( "Question 465377: If 3x^2-2x+7=0, then (x-1/3)^2=?? \n" ); document.write( "
Algebra.Com's Answer #318904 by stanbon(75887)\"\" \"About 
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If 3x^2-2x+7=0
\n" ); document.write( "Solve for \"x\":
\n" ); document.write( "x = [2+-sqrt(4-4*3*7)]/6
\n" ); document.write( "---
\n" ); document.write( "x = [2 +- sqrt(80)]/6
\n" ); document.write( "---
\n" ); document.write( "x = [2 +- 4sqrt(3)]/6
\n" ); document.write( "----
\n" ); document.write( "x = [1 + 2sqrt(3)]/3 or x = [1 - 2sqrt(3)]/3
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\n" ); document.write( " then (x-1/3)^2=??
\n" ); document.write( "---
\n" ); document.write( "If x = [1+2sqrt(3)]/3
\n" ); document.write( "then (x-(1/3)^2
\n" ); document.write( "= [[1+2sqrt(3)]/3 - (1/3)]^2
\n" ); document.write( "= [2sqrt(3)/3]^2
\n" ); document.write( "= 4/3
\n" ); document.write( "----
\n" ); document.write( "If x = [1-2sqrt(3)]/3
\n" ); document.write( "then (x-(1/3)^2 = 4/3
\n" ); document.write( "==========================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
\n" ); document.write( "
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