How do you solve #(x + 3) ^ { 2} - 10= 2#?

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Answer 1 · 2017-08-08T19:58:48

#x = +sqrt12-3 " " or x = -sqrt12-3#

#x= 0.464" "or " "x = -6.464#

It is not necessary to square the bracket. Isolate it on the left and then find the square root of both sides.

#(x+3)^2 = 2+10#

#(x+3)^2 = 12" "larr# find the square root

#x+3 = +-sqrt12#

#x = +sqrt12-3 " " or x = -sqrt12-3#

#x= 0.464" "or x = -6.464#