NP-completeness and NP problems
Recall a few definitions. A language $L \subseteq \{0,1\}^*$ is polynomial-time reducible to a language $L' \subseteq \{0,1\}^*$, often denoted as $L \leq_p L'$, if there is a polynomial-time computable function $f : \{0,1\}^* \to \{0,1\}^*$ such that for every $x \in \{0,1\}^*$, $x \in L$ iff $f(x) \in L'$. Now, $L'$ is $\text{NP}$-hard if $L \leq_p L'$ for every $L \in \text{NP}$. $L'$ is $\text{NP}$-complete if $L'$ is $\text{NP}$-hard and $L' \in \text{NP}$.

It is easy to see that if $L$ is $\text{NP}$-hard and $L \in \text{P}$, then $\text{P} = \text{NP}$. Likewise, if $L$ is $\text{NP}$-complete, then $L \in \text{P}$ iff $\text{P} = \text{NP}$. The intuitive reason is that if $f$ and $g$ are functions that grow at most as $n^c$ and $n^d$ respectively, then the composition $f(g(n))$ grows at most $n^{cd}$, which is also polynomial.

Assume you found a polynomial-time algorithm $A$ for a $\text{NP}$-complete problem $X$. Now, you could in polynomial time transform another $\text{NP}$-complete problem to $X$ and solve it using $A$. This is why we often say that it suffices to study whether any $\text{NP}$-complete problem can be decided in polynomial time.