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Math Help - How to solve this equation

  1. #1
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    How to solve this equation

    I need step by step help on solving this equation:

    4x^2 - 15x = 3
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  2. #2
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    \displaystyle 4x^2 - 15x = 3

    \displaystyle 4x^2 - 15x -3= 0


    by the quadartic formula

    \displaystyle ax^2 + bx +c= 0 \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

    \displaystyle 4x^2 - 15x -3= 0 \implies x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}

    You got it from here?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    \displaystyle 4x^2 - 15x = 3

    \displaystyle 4x^2 - 15x -3= 0


    by the quadartic formula

    \displaystyle ax^2 + bx +c= 0 \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

    \displaystyle 4x^2 - 15x -3= 0 \implies x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}

    You got it from here?
    I'm confused on what to do next ...
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  4. #4
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    given the quadratic formula with all variables filled in, you simply simplify down (notice the pun there)

    x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}

    x = \frac{15\pm\sqrt{225-48}}{8}

    x = \frac{15\pm\sqrt{177}}{8}

    Using order of operations, of course.
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  5. #5
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    Quote Originally Posted by fireballs619 View Post
    given the quadratic formula with all variables filled in, you simply simplify down (notice the pun there)

    x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}

    x = \frac{15\pm\sqrt{225-48}}{8}

    x = \frac{15\pm\sqrt{177}}{8}

    Using order of operations, of course.
    thanks!
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