When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are \[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\]
\[ \displaylines{ \left(x+y\right)^2 &=\left(x+y\right)\left(x+y\right) \\ &=x^2+2xy+y^2 } \]
\[x^{5}-x^{4}-1=x^{5}-x^{4}+x^{3}-x^{3}-1\]
\[\frac{n!}{k!(n-k)!} = \binom{n}{k}\]