Quadratic 3

I will lift the formula from Vieta's formulas page on Wikipedia that $\sum_{1\le i_1 < i_2 < \cdots < i_k\le n} x_{i_1}x_{i_2}\cdots x_{i_k}=(-1)^k\frac{a_{n-k}}{a_n}$ for a polynomial of $P(x)=a_nx^n + a_{n-1}x^{n-1} +\cdots + a_1 x+ a_0$ . When matching up with this equation, it will be $x_1+x_2\ = (-1)^1(-\frac{94}{55}) = \frac{94}{55}$ My application on the above solution is a little bit reverse to it, as I observe that the sign will be reversed from the sum of odd-multiple and even-multiple terms. In my study, it can be concluded that if $a_nx^n$ is the n-th term from the left of polynomial with degrees of each monomial sorted in descending order, the sign of $(-1)^k$ in the equation above will equal to the sign of $(-1)^{1+n}$ . Here, when this polynomial are sorted in descending order, the sum we are looking for, the $x_1+y_1$ is the coefficient in the second position from left, hence $(-1)^{1+2}$