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Math Help - Solving a quadratic equation

  1. #1
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    Solving a quadratic equation

    I have the following quadratic equation:
    \frac{1}{2}kd^2-\mu_kmgl-\frac{1}{2}mv_B^2=0

    I would like to solve for d.
    So I wanna use this formula:
    x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}

    Now I know that here I have:
    a=\frac{1}{2}k

    However is it true to say that:
    b=0
    c=-\mu_kmgl-\frac{1}{2}mv_B^2

    ???? I'm wondering if c can include operators like plus and minus?
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  2. #2
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    Re: Solving a quadratic equation

    Yes, that is correct.
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  3. #3
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    Re: Solving a quadratic equation

    yes it can
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