If $321_{b}$ is equal to the base 10 integer 57, find $b$ given that $b>0$.
Converting $321_{b}$ to base 10 and setting it equal to 57, we find that  \begin{align*} 3(b^2)+2(b^1)+1(b^0)&=57
\\ 3b^2+2b+1&=57
\\\Rightarrow\qquad 3b^2+2b-56&=0
\\\Rightarrow\qquad (3b+14)(b-4)&=0
\end{align*}This tells us that $b$ is either $-\frac{14}{3}$ or $4$. We know that $b>0$, so $b=\boxed{4}$.