To supplement my previous answer, you should think about what it means economically for a cost function's derivative to have a root.
Mathematically, it means that aggregate costs are falling at quantities less than that root. Let's test that.
[MATH]c(10) = 250 + \dfrac{10^2 - 62(10)}{100} = 250 - \dfrac{520}{100} = 244.8.[/MATH]
[MATH]c(20) = 250 + \dfrac{20^2 - 62(20)}{100} = 250 - \dfrac{-840}{100} = 241.6 < 244.8.[/MATH]
Let's look at that first derivative you calculated, the marginal cost.
[MATH]c'(x) = \dfrac{x - 31}{50} \implies\\
c'(10) = \dfrac{10 - 31}{100} < 0 > \dfrac{20 - 31}{100} = c'(20).[/MATH]But economically, at least if the producer is a price taker rather than a price maker, then, at any level of production where marginal cost is negative (meaning total cost is falling) the producer necessarily increases revenue by making and selling more product while actually reducing costs. Expanding production when marginal costs are negative necessarily increases the profits of a price taker: revenues go up while expenses go down. Thus, a price taker would be stupid as a stone not to expand production AT LEAST to where marginal cost is zero. In fact there is a fundamental theorem in the theory of the firm that, to maximize profits, a producer should increase production to the point where total costs are increasing and also the marginal cost equals price. (That theorem has implications for a price maker that do not apply to a price taker.)
So the number that you calculated has economic significance, but I do not think it was what was being looked for.
