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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).
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