How do you solve #x^2+4x-3=0# by completing the square?

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Answer 1 · 2015-05-11T17:47:05

#0 = x^2+4x-3 = x^2+4x+4-7 = (x+2)^2-7#

Adding 7 to both ends we get:

#(x+2)^2 = 7#

So #x + 2 = +-sqrt(7)#

Subtracting 2 from both sides, we arrive at:

#x = +-sqrt(7)-2#.

In general, #ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))#.