Given that,
$x^{2}+2x+6=0$
now as r is root of the equation , so we can write
$r^{2}+2r+6=0$
or , $r^{2}=-2r-6$
we have to find
(r+2)(r+3)(r+4)(r+5) [ where r is the root of the equation ]
= $(r^{2}+5r+6)(r^{2}+9r+20)$
= ($-2r-6 +5r+6$)($-2r-6 +9r+20$)
=$3r (7r+14)$
=$21(r^{2}+2r)$ [$(r^{2}+2r)=-6$]
=21*(-6)
= -126
answer is (-126).