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Math Help - Need a way to solve this equation

  1. #1
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    Need a way to solve this equation

    Hi there guys, how are you all?

    I really need to find a way to solve this equation, but I have forgotten most of my maths, i just need a lead on the method of solving this equation.


    0=(-6.5x10^-14*<i>X^3</i>)-(1x10^-16*<i>X^2</i>)-(<i>X</i>/1x10^-8)+(1.3036x10^32)
    capital X is the variable. I need to solve and get the equivalent of the X. Please someone guide me on the method. Thanks very much! cheers!
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by thavamaran View Post
    Hi there guys, how are you all?

    I really need to find a way to solve this equation, but I have forgotten most of my maths, i just need a lead on the method of solving this equation.


    0=(-6.5x10^-14*<i>X^3</i>)-(1x10^-16*<i>X^2</i>)-(<i>X</i>/1x10^-8)+(1.3036x10^32)
    capital X is the variable. I need to solve and get the equivalent of the X. Please someone guide me on the method. Thanks very much! cheers!
    please check your posting ... is that what you have wrote there this :

     0=(-6.5\cdot 10^{-14} \cdot X^3)-(1\cdot 10^{-16}\cdot X^2)-(\frac {X}{1\cdot 10^{-8}})+(1.3036\cdot 10^{32})
    Last edited by yeKciM; September 25th 2010 at 09:25 AM.
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  3. #3
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    Quote Originally Posted by thavamaran View Post
    Hi there guys, how are you all?

    I really need to find a way to solve this equation, but I have forgotten most of my maths, i just need a lead on the method of solving this equation.


    0=(-6.5x10^-14*<i>X^3</i>)-(1x10^-16*<i>X^2</i>)-(<i>X</i>/1x10^-8)+(1.3036x10^32)
    capital X is the variable. I need to solve and get the equivalent of the X. Please someone guide me on the method. Thanks very much! cheers!
    Use technology (graphics or CAS calculator, computer etc.) or a numerical method to get a solution to the desired accuracy.
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  4. #4
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    Hi yeKciM, yes, that is my question, thank you!

    And sorry for not following forum rules! Wont happen again.
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  5. #5
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    Quote Originally Posted by thavamaran View Post
    Hi yeKciM, yes, that is my question, thank you!

    [snip]
    See post #3.
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