How do you solve the quadratic equation by completing the square: #x^2+6x-4=0#?

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Answer 1 · 2015-07-21T09:51:56

Add #13# to both sides to make the left hand side a perfect square trinomial, then take the square root of both sides and subtract #3# to find:

#x = -3+-sqrt(13)#

Add #13# to both sides to get:

#13 = x^2+6x+9 = (x+3)^2#

#(x+3)^2=13#

Take the square root of both sides to get rid of the exponent:

#rarrsqrt((x+3)^2)=+-sqrt13#

So #x+3 = +-sqrt(13)#

Subtract #3# from both sides to get:

#x = -3+-sqrt(13)#