How do you solve #x^2 + 4x = 6# by completing the square?

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Answer 1 · 2018-03-31T12:15:22

Solution: #x=-2+ sqrt10 and x= -2- sqrt10#

#x^2+4x=6 or x^2+4x+4=6+4 # or

#(x+2)^2=10 or x+2 =+- sqrt 10# or

#x=-2+- sqrt10#

#:.x=-2+ sqrt10 and x= -2- sqrt10#

Solution: #x=-2+ sqrt10 and x= -2- sqrt10# [Ans]

Answer 2 · 2018-03-31T12:55:29

#x^2+4x=6#

Adding 4 on both sides.

#=>x^2+4x +4 =6 + 4#

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

#=> x+2 = +-sqrt10#

# => x = -2 +- sqrt10#