The most common version for Vieta's formulas is the one we use for quadratics. If we have a quadratic polynomial ax^2 + bx + c = 0, then the (two) roots $x_1$, and $x_2$ satisfy $x_1 + x_2 = -\dfrac{b}{a}$, and $x_1x_2 = \dfrac{c}{a}$. 

We can extend this to a more general version for polynomials of higher degree (which is what the problem is using). 

If we have a polynomial $$a_{n}x^{n} + a_{n-1} x^{n-1} + \cdots + a_1x + a_0,$$ with roots $x_1, \dots x_n$ then Vieta's formulas say that $$\begin{aligned} x_1 + x_2 + \cdots + x_n &= -\dfrac{a_{n-1}}{a_n} \\ x_1x_2 + x_1x_3 + \cdots + x_1x_n + x_2x_3 + \cdots + x_2x_n + \cdots + x_{n-1}x_n &= \dfrac{a_{n-2}}{a_n} \\ &\vdots \\ x_1x_2x_3\cdots x_n &= (-1)^n \dfrac{a_0}{a_n} \end{aligned}$$

In this case we have a degree $5$ polynomial, and we want to calculate $\sum_{i=1}^{5} x_i = x_1 + \cdots x_5$, so Vieta's formulas tell us that sum is equal to $- \dfrac{a_4}{a_5}$. For the polynomial in question $a_4 = 0$ and $a_5 = 1$, so the sum of the roots is equal to $0$.