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

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

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Answer 1 · 2016-04-10T11:14:48

$x = - 5; x = 1$ $x=-5 ; x=1$

Explanation:

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

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

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How do you solve x2+4x−5=0x^2+4x-5=0 by completing the square?

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Explanation:

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