A problem $\Pi$ is $\textrm{NP}$-hard if every problem in $\Pi' \in \textrm{NP}$ can be reduced to $\Pi$ via a Karp reduction $\le_p$, i.e., if $\Pi' \le_p \Pi$.
A problem $\Pi$ is $\textrm{NP}$-complete if it is $\textrm{NP}$-hard and belongs to $NP$.
Intuitively, a $\textrm{NP}$-complete problem is as hard as the hardest problem in $\textrm{NP}$ while a $\textrm{NP}$-hard problem might be even harder.
Finding a polynomial-time algorithm $A$ for any $\textrm{NP}$-hard problem $\Pi$ (therefore also for any $\textrm{NP}$-complete problem) would immediately yield a polynomial-time algorithm for all problems in $\textrm{NP}$: reduce (in polynomial-time) an instance $x$ your problem $\Pi' \in \textrm{NP}$ of choice to an instance $y$ of $\Pi$ and then solve $y$ it using $A$.