Solve x^2+4x-5=0 by completing the square.

x^2+4x-5=0(x+2)^2-(2)^2-5=0subtract 2^2 as (x+2)^2 = x^2+4x+4 when we want x^2+4x(x+2)^2-4-5=0(x+2)^2-9=0(x+2)^2=9square root both sidesx+2=+/-3x=-2+/-3x=-2+3 or x=-2-3x=1 or x=-5

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Answered by Evie M. Maths tutor

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