\(\displaystyle ab+2a+2b=-16, \\ \text{so}\\ab+2a+2b+4=-12, \\ (a+2)(b+2)=-12 \dots \dots (1)\)
Similarly,
\(\displaystyle (a+2)(c+2)=30 \dots \dots(2) \\ (b+2)(c+2)=-10 \dots \dots(3).\)
Let
\(\displaystyle a+2=t,\\ \text{then from (1) and (2), }\\b+2=-12/t \\ c+2=30/t.\)
Substitute into (3),
\(\displaystyle (-12/t)(30/t)=-10,\\ t^{2}=36,\\ t = \pm6.\)
\(\displaystyle \text{If }\quad t=6,\text{ then }\\a=4,\quad b=-4, \quad c=3.\)
\(\displaystyle \text{If }\quad t=-6, \text{then} \\a=-8, \quad b=0,\quad c=-7.\)
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