How do you solve #w^2+4w-2=0# by completing the square?

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Answer 1 · 2015-05-23T22:18:28

#0 = w^2+4w-2 = w^2+4w+4-4-2 = (w+2)^2-6#

Add #6# to both ends to get:

#(w+2)^2=6#

So

#w+2 = +-sqrt(6)#

Subtract #2# from both sides to get:

#w=-2+-sqrt(6)#