The equation $x^3 - 4x^2 + 5x - \frac{19}{10} = 0$ has real roots $r,$ $s,$ and $t.$  Find the length of the long diagonal of a box with sides of length $r,$ $s,$ and $t.$
To find the length of the long diagonal of a box, I need to use the Pythagorean theorem in three dimensions. That means I need to square the lengths of the three edges that meet at a corner of the box, add them up, and take the square root. So the long diagonal is $\sqrt{r^2 + s^2 + t^2}.$ Now I just need to find $r^2 + s^2 + t^2.$ One way to do that is to use the fact that $r,$ $s,$ and $t$ are the roots of the given equation, and apply Vieta's formulas. Vieta's formulas relate the coefficients of a polynomial to the sums and products of its roots. For a cubic equation like this one, Vieta's formulas say that $r + s + t = 4,$ $rs + rt + st = 5,$ and $rst = \frac{19}{10}.$ To get $r^2 + s^2 + t^2,$ I can use the identity $(r + s + t)^2 = r^2 + s^2 + t^2 + 2(rs + rt + st).$ Rearranging this identity, I get $r^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + rt + st).$ Plugging in the values from Vieta's formulas, I get $r^2 + s^2 + t^2 = 4^2 - 2(5) = 6.$ Therefore, the long diagonal of the box is $\sqrt{6}.$ 