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solve:
(x+2)^2=12
(x+2)^2 = 12
x^2+4x+4-12 = 0
x^2+4x-8 = 0
use the formula x = -b+-sq rt b^2-4ac/2a
where a =1 b = 4 c= -8
substitute the values of a,b&c
x = -4+-sq rt 4^2-4.1.(-8)/2.1
= -4+-sq rt 16+32/2
= -4+sq rt 48/2
= -4+-4sq rt3/2
= 2(-2+-sq rt3)/2
= -2+-sq rt3