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Math Help - Quadratic equation

  1. #1
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    Question Quadratic equation

    trying to figure out how my teacher got this answer using the quadratic equation.

    Problem: 3x^2 + 2x -6

    This is how he did it in class

    3x^2 + 2x - 6 = 0
    x= -2 +- square root 4-4(3)(-6) / 2(3)

    x= -2 +- square root 4 * 19 / 6

    = -1/3 + square root 19/3

    = square root 19-4 / 3

    That is the final answer but I dont know how he got that.

    Can anyone please help
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  2. #2
    GAMMA Mathematics
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    3x^2 + 2x -6 = 0

    Use the quadratic equation:
    x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

    x=\frac{-2 \pm \sqrt{4-4(-6)(3)}}{6}
    x=\frac{-2 \pm \sqrt{76}}{6}
    x=\frac{-2 \pm \sqrt{4*19}}{6}
    x=\frac{-2 \pm 2\sqrt{19}}{6}
    x=\frac{-1 \pm \sqrt{19}}{3}
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  3. #3
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    ok so using the same method

    how does this quadratic work

    3x^2 + 2x -6.5

    x= -2 +- square root 4-4(3)(-6.5) / 2 (3)

    x = -2 +- square root 82 / 6

    then I get stuck right here.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by winterjm View Post
    ok so using the same method

    how does this quadratic work

    3x^2 + 2x -6.5

    x= -2 +- square root 4-4(3)(-6.5) / 2 (3)

    x = -2 +- square root 82 / 6

    then I get stuck right here.
    It would help, for starters, if you would use parenthesis. Your last line should read:
    x = (-2 +- square root(82))/6

    Right-o then. Since \sqrt{82} can't be simplified you are done. There is nothing left to do for this one.

    -Dan
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