{tex}{1\\over x}-{1\\over x-2}=3{/tex}\xa0{tex}{x-2-x\\over x(x-2)}=3{/tex} [by taking L.C.M.]-2=3×(x(x-2))-2=3(x2-2x)-2=3×2-6x3x2-6x+2=0On comparing with ax2+bx+c=0 we get\xa0a=3 b= -6 c=2By using quadratic formula\xa0{tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}{tex}x = {-(-6) \\pm \\sqrt{(-6)^2-4×3×2} \\over 2×3}{/tex}{tex}x = {6 \\pm \\sqrt{36-24} \\over 6}{/tex}{tex}x = {6 \\pm \\sqrt{12} \\over 6}{/tex}{tex}x = {6 \\pm 2\\sqrt{3} \\over 6}{/tex}{tex}x = {2(3\\pm \\sqrt{3}) \\over 6}{/tex}{tex}x = {3\\pm \\sqrt{3} \\over 3}{/tex}So roots are\xa0{tex}{3 +\\sqrt{3} \\over 3}{/tex} and {tex}{3 – \\sqrt{3} \\over 3}{/tex}