The correct option is A. [tex]x^{2} -x-10=0[/tex].
Given quadratic equation [tex]\frac{2}{3x-6} -\frac{1}{3x+6} =\frac{1}{3}[/tex]
On solving, [tex]\frac{2(3x+6)-1(3x-6)}{(3x-6)(3x+6)} =\frac{1}{3}[/tex]
Further, [tex]\frac{6x+12-3x+6}{9x^{2} -6^{2} } =\frac{1}{3}[/tex]
[tex]\frac{3x+18}{9x^{2} -36} =\frac{1}{3}[/tex]
Taking 3 common from numerator and 9 common from denominator, we have
[tex]\frac{3(x+6)}{9(x^{2} -4)} =\frac{1}{3}[/tex]
[tex]\frac{(x+6)}{3(x^{2} -4)} =\frac{1}{3}[/tex]
[tex]\frac{x+6}{x^{2} -4} =1[/tex]
[tex]x+6=x^{2} -4[/tex]
On rearranging in decreasing powers of x we get,[tex]x^{2} -x-10=0[/tex].
Hence on solving, We get [tex]x^{2} -x-10=0.[/tex]
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