How do you solve $x^2+4x-5=0$ by completing the square?

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Answer 1 · 2016-04-10T11:14:48

$x=-5 ; x=1$

$x^2+4x-5 =0 or x^2+4x+4-9=0 or(x+2)^2-3^2=0 or(x+2+3)(x+2-3)=0 or(x+5)(x-1)=0$ So either $x+5=0 ; x=-5 or x-1=0 ; x=1$ [Ans]

How do you solve x^2+4x-5=0x^2+4x-5=0 by completing the square?