This is a $4^{th}$ degree equation, therefore totally $4$ roots exist. By the Vieta's formula, the sum of these roots equals the coefficient of the $x^3$ term which is $0$. Hence, the $4^{th}$ root must be $-1$. From here, it is possible to determine all the $a$, $b$, and $c$. A convenient way is write the given equation as $$x^4 + ax^2 + bx+c= (x-2)(x+3)(x-5)(x+4)$$ because these four numbers are its root. Setting $x=1$ to the above relation yields $$1+a+b+c=(-1)\cdot 4 \cdot(-4)\cdot 5\implies a+b+c=\boxed{79}$$