We have 2×2 – 7x + 3 = 0{tex}\\implies2( x^2 – {7 \\over 2}x + {3\\over 2}) = 0{/tex}{tex}\\implies\u200b\u200b x^2 – {7 \\over 2}x + {49 \\over 16} = {-3 \\over 2} +{ 49 \\over 16}{/tex} (Adding 49/16 to both sides){tex}\\implies x^2 -2 \\times x \\times {7 \\over 4} + ({7 \\over 4})^2 = {-24 +49 \\over 16}{/tex}{tex}\\implies (x-{7\\over4})^2 = {25 \\over 16}{/tex}{tex}\\implies x-{7\\over 4}= \\pm \\sqrt({25 \\over 16}){/tex}{tex}\\implies x={7\\over 4} \\pm {5 \\over 4}{/tex}{tex}\\implies x={7\\over 4} + {5 \\over 4}\\, and \\,x={7\\over 4} – {5 \\over 4}{/tex}{tex}\\implies x=3\\, and \\,{1\\over 2}{/tex}{tex}\\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\\over 2{/tex}.