How to calculate the enthalpy change for the formation of ethylene epoxide from ethylene? I get everything up until the very end. We need to identify the bond enthalpies first:
\begin{align}
\ce{C=C} &amp;= 612\\
\ce{O=O} &amp;= 248\\
\text{SUM of these}&amp;= 860\\
\end{align}

\begin{align}
\ce{C-C} &amp;= 348\\
2\cdot(\ce{C-O}) &amp;= 720\\
\text{SUM of these} &amp;= 1068\\
\end{align}
We know that enthalpy change is = products - reactants, which is $1068 - 860 = 208$, but the problem comes here: I know a minus sign is supposed to go there, but
how do we know that? Exothermic reaction? Is it because of the oxygen, suggesting it's a combustion reaction or not?
When you use bond enthalpies to calculate $\Delta H$, you need to subtract the sum of the bond enthalpies for the products from the sum of the bond enthalpies for the reactants.  In other words "reactants - products".  This is because all bond enthalpies are positive, since energy must be put into the bond to break it.  Bond formation releases energy; the minus sign compensates for this.  In short, bonds of reactant molecule must break, and new are formed to form product molecules.  So, your answer should be negative.