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solve:
(x+2)^2=3
consider (x+2)^2 = 3
x^2+4x+4-3= 0
x^2+4x+1 = 0
use the formula x = -b +-sq rt b^2-4ac/2a where a = 1 b = 4 c = 1
x = -1+-sq rt 1^2-4.1.1/2.1
= -1+-sq rt 1-4/2
= -1+sq rt-3/2 since sq rt of negative number cannot be found we consider the solution as x = (-1+-sqrt-3)/2