1 answer

How do you solve #y^2 + 12y + 11 = 0#?

Question:

How do you solve #y^2 + 12y + 11 = 0#?

Answers

#y=-1 or y=-11#

Explanation:

#y=-6+-sqrt(6^2-11)#
#y=-6+-sqrt(25)#
#y=-6+-5#
#y=-1 or y=-11#

.

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