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Math Help - Test

  1. #1
    Newbie
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    Feb 2011
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    Test

    \begin{array}{l}<br />
\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\<br />
\frac{{n!}}{{r!\left( {n - r} \right)!}}\\<br />
\frac{1}{2}<br />
\end{array}
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by robertog View Post
    \begin{array}{l}<br />
\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\<br />
\frac{{n!}}{{r!\left( {n - r} \right)!}}\\<br />
\frac{1}{2}<br />
\end{array}
    You might find centred {c} looks better than left justified {l} for this. Also if you start with \dfrac you will get larger fractions, and it would benefit from blank lines between the equations

    \displaystyle \begin{array}{c}<br />
\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ \\<br />
\dfrac{{n!}}{{r!\left( {n - r} \right)!}}\\ \\<br />
\dfrac{1}{2}<br />
\end{array}
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  3. #3
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    China
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    b^2 + 4ac
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