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Math Help - quadratic solution

  1. #1
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    quadratic solution

    i've got 3x^2-6xL+2L^2 = 0

    I want to get x = 0.423L

    How do I get to there? Clear explanation needed.. ty
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  2. #2
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    Quote Originally Posted by custer View Post
    i've got 3x^2-6xL+2L^2 = 0

    I want to get x = 0.423L

    How do I get to there? Clear explanation needed.. ty
    Assuming L is a constant you can use the quadratic formula. Because L is a constant it can be incorporated into ax^2+bx+c=0

    In your case:

    a = 3
    b = -6L
    c = 2L^2

    The plug those into the quadratic formula to find x in terms of L:

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{6L \pm \sqrt{36L^2 - 24L^2}}{6}

    simplify for L

    One of the solutions is \frac{3-\sqrt3}{3} L = 0.423L \text { (3dp) }

    There will be another solution though because of the plus/minus
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