How do you solve #3x^2+14x=5# by completing the square?

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Answer 1 · 2015-05-12T19:08:16

Dividing through by #3#, we get:

#x^2+(14/3)x = 5/3#

Now try squaring #x+7/3#

#(x+7/3)^2 = (x+7/3)(x+7/3)#

#= x^2 + (14/3)x + 49/9#

#= 5/3 + 49/9#

#= 15/9+49/9#

#= 64/9#

#= (8^2)/(3^2)#

Hence #x + 7/3 = +-8/3#

Subtracting 7/3 from both sides, we get

#x = +-8/3 - 7/3#

So #x = 1/3# or #x = -15/3 = -5#.