Question

Solve this by using Formula[tex]\tt 4x^2-2x-1=0[\tex]​

Answer 1

$\huge {\boxed{\boxed {\color {red}{\mathfrak {Answer}}}}}$

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

Answer 2

$\large\bold{\underline{\underline{Answer:-}}}$

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