Test

**robertog** · February 24th 2011, 03:09 PM

$\begin{array}{l} \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ \frac{{n!}}{{r!\left( {n - r} \right)!}}\\ \frac{1}{2} \end{array}$

**CaptainBlack** · February 24th 2011, 10:04 PM

Originally Posted by robertog

$\begin{array}{l} \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ \frac{{n!}}{{r!\left( {n - r} \right)!}}\\ \frac{1}{2} \end{array}$

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

$\displaystyle \begin{array}{c} \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ \\ \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\\ \\ \dfrac{1}{2} \end{array}$

**wsc810** · March 10th 2011, 02:58 PM

$b^2 + 4ac$

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