How do you solve by completing the square: #7w² + 6w + 7 = 0#?

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Answer 1 · 2015-03-30T10:45:51

In this way:

#7w^2+6w+7=0rArr7(w^2+6/7w)+7=0rArr#

#7(w^2+6/7w+(3/7)^2-(3/7)^2)+7=0rArr#

#7(w^2+6/7w+(3/7)^2)-7*9/49+7=0rArr#

#7(w+3/7)^2-9/7+7=0rArr#

#7(w+3/7)^2=9/7-7rArr#

#7(w+3/7)^2=(9-63)/7rArr#

#7(w+3/7)^2=-54/7#

that is an impossible equation, because a square can't be negative!