How do you solve #r^ { 2} + 16r + 45= 0#?

1 Answer
Binayaka C.
Aug 8, 2017

Solution : #r = -8+ sqrt19 , r = -8- sqrt19#

Explanation:

#r^2+16r+45=0 or r^2+16r+64 -64+45 # or

# (r+8)^2 -19=0 or (r+8)^2 =19 or (r+8) =+- sqrt19# or

#r = -8+- sqrt19 #

Solution : #r = -8+ sqrt19 , r = -8- sqrt19# [Ans]

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