Why is delta-H negative here?
The bond enthalpy is conventionally the enthalpy change of bond breaking, not of bond forming, in the contrary to formation enthalpy of compounds.
This way they are all positive.
The consequence is, the order of products and reactants in computation the enthalpy difference has to be switched.
So if the summary bond enthalpy of products is higher than of reactants, the difference of enthalpy as the reaction enthalpy is negative, not positive, and the reaction is exothermic.
Formation enthalpies:
$$\Delta H_\mathrm{r}=\Sigma  \left( \Delta H_\mathrm{f, products} \right) - \Sigma \left( \Delta H_\mathrm{f, reactants} \right)   $$
Bond enthalpies:
$$\Delta H_\mathrm{r}=  \Sigma   \left(  \Delta H_\mathrm{b, reactants} \right)   - \Sigma  \left(  \Delta H_\mathrm{b, products}  \right) $$