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