Use Simon's favorite factoring trick; it's going to make it so much easier:

\((a+2)(b+2)=-12\\ (a+2)(c+2)=30\\ (b+2)(c+2)=-10\)

Multiply all the equations to get:

\(((a+2)(b+2)(c+2))^2=3600\\ (a+2)(b+2)(c+2)= \pm 60\)

In the first case where the right-hand side of the equation above is positive, to solve for a, divide \((a+2)(b+2)(c+2)=60\) by \((b+2)(c+2)=-10\) to get \(a+2=-6 \rightarrow a=-8\). Do that the same with b and c to get \(b = 0\) and \(c=-7\).

In the second case where the right-hand side of the equation is negative, do the exact same thing mentioned above to get \(a = 4, b = -4, c = 3\).

Therefore, the solutions to this system of the equation are:

\(\boxed{a = -8, b = 0, c = -7 \text{ or } a = 4, b = -4, c = 3} \)

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