Question

Answer for the 25th question​

Answer 1

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$\tt 4x^2-2x-1=0 \\ \\ \tt Here,a=4,b=2\:and\:c=-1 \\ \\ \tt Using,quadratic\: formula\:, we\: get \\ \\ \tt x=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a} \\ \\ \tt \implies =\dfrac {-2\pm \sqrt {2^2-4×4×-1}}{2×4} \\ \\ \tt \implies x=\dfrac {-2\pm \sqrt {4+16}}{8} \\ \\ \tt \implies =\dfrac {-2\pm \sqrt {20}}{8} \\ \\ \tt \implies =\dfrac {-2\pm 2 \sqrt5}{8} \\ \\ \tt \implies x=\dfrac {-2+2\sqrt5}{8}\:or\: x=\dfrac {-2-2\sqrt5}{8}$

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