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

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