Quinquennial Sum

The problem can be solved using Newton's Sum method.

Let the roots of the equation be $a$ , $b$ , $c$ and $d$ , then:

$\begin{cases} S_1 = a+b+c+d & = 0 \\ S_2 = ab+ac+ad+bc+bd+cd & = -7 \\ S_3 = abc+abd+acd+bcd & = -4 \\ S_4 = abcd & = -3 \end{cases}$

And

$\begin{cases} P_1 = a+b+c+d & = S_1 & = 0 \\ P_2 = a^2+b^2+c^2+d^2 & = S_1P_1-2S_2 & = 14 \\ P_3 = a^3+b^3+c^3+d^3 & = S_1P_2-S_2P_1 + 3S_3 & = -12 \\ P_4 = a^4+b^4+c^4+d^4 & = S_1P_3-S_2P_2 + S_3P_1 - 4S_4 & = 110 \\ P_4 = a^5+b^5+c^5+d^5 & = S_1P_4-S_2P_3 + S_3P_2 - S_4P_1 & = \boxed{-140} \end{cases}$

The calculations can be easily done with a spreadsheet as follows: