What is the minimum value of $5x^2-20x+1357$?
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 $\left(\dfrac{4}{2}\right)^2=4$ 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 $\left(x-2\right)^2\ge0$, the minimum value is when the squared term is equal to $0$.  So the minimum value is $y=5\left(x-2\right)^2+1337=5\cdot0+1337=\boxed{1337}$.
1337
Level 4
Algebra
Problem: What is the minimum value of $5x^2-20x+1357$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum value of the quadratic function $5x^2 - 20x + 1357$, we recognize that it is a parabola that opens upwards since the coefficient of $x^2$ is positive. The minimum value occurs at the vertex of the parabola. 

The x-coordinate of the vertex can be found using the formula $x = -\frac{b}{2a}$, where $a = 5$ and $b = -20$. 

Calculating this gives:
x = -(-20) / (2 * 5) = 20 / 10 = 2.

Next, we substitute $x = 2$ back into the original quadratic function to find the minimum value:
5(2)^2 - 20(2) + 1357.

Calculating each term:
5(4) = 20,
-20(2) = -40,
Thus, we have:
20 - 40 + 1357 = -20 + 1357 = 1337.

Therefore, the minimum value of the function is $1337$.<|END_OF_LATENT|> The final answer is $1337$.