Find all solutions to

$\displaystyle ab+2a+2b=-16, \\ \text{so}\\ab+2a+2b+4=-12, \\ (a+2)(b+2)=-12 \dots \dots (1)$

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),

$\displaystyle (-12/t)(30/t)=-10,\\ t^{2}=36,\\ t = \pm6.$

 $\displaystyle \text{If }\quad t=6,\text{ then }\\a=4,\quad b=-4, \quad c=3.$

$\displaystyle \text{If }\quad t=-6, \text{then} \\a=-8, \quad b=0,\quad c=-7.$