jmc

Let $y=5x^2-20x+1357$. First, complete the square as follows: $y=5x^2-20x+1357=5(x^2-4x)+1357$. To complete the square, we need to add ${\left(\frac{4}{2}\right)}^2=4$ after the $-4x$. So we have $y+20=5\left(x^2-4x+4\right)+1357$. This gives $y=5{\left(x-2\right)}^2+1337$.

Now, since ${\left(x-2\right)}^2\ge 0$, the minimum value is when the squared term is equal to $0$. So the minimum value is $y=5{\left(x-2\right)}^2+1337=5\cdot 0+1337=\boxed{1337}$.