The side lengths of a scalene triangle are roots of the polynomial $$x^3-20x^2+131x-281.3.$$ Find the square of the area of the triangle.
1. Let the side lengths of the scalene triangle be \(a\), \(b\), and \(c\). These are the roots of the polynomial \(x^3 - 20x^2 + 131x - 281.3\).

2. By Vieta's formulas, we know:
   \[
   a + b + c = 20,
   \]
   \[
   ab + bc + ca = 131,
   \]
   \[
   abc = 281.3.
   \]

3. To find the square of the area of the triangle, we use Heron's formula. First, we compute the semi-perimeter \(s\):
   \[
   s = \frac{a + b + c}{2} = \frac{20}{2} = 10.
   \]

4. Heron's formula for the area \(A\) of a triangle is:
   \[
   A = \sqrt{s(s-a)(s-b)(s-c)}.
   \]
   We need to find \(s(s-a)(s-b)(s-c)\):
   \[
   s(s-a)(s-b)(s-c) = 10 \cdot (10-a) \cdot (10-b) \cdot (10-c).
   \]

5. We expand \((10-a)(10-b)(10-c)\):
   \[
   (10-a)(10-b)(10-c) = 1000 - 100(a+b+c) + 10(ab+bc+ca) - abc.
   \]

6. Substitute the values from Vieta's formulas:
   \[
   (10-a)(10-b)(10-c) = 1000 - 100 \cdot 20 + 10 \cdot 131 - 281.3.
   \]

7. Simplify the expression:
   \[
   (10-a)(10-b)(10-c) = 1000 - 2000 + 1310 - 281.3 = 28.7.
   \]

8. Therefore:
   \[
   s(s-a)(s-b)(s-c) = 10 \cdot 28.7 = 287.
   \]

9. The square of the area of the triangle is:
   \[
   A^2 = s(s-a)(s-b)(s-c) = 287.
   \]

The final answer is \(\boxed{287}\).