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

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Answer 1 · 2015-04-30T00:11:57

$4x^2 - 2x - 5 = 4x^2 - 2x + 1/4 - 5 - 1/4 =$
$(2x - 1/2)^2 = 5 + 1/4 = 21/4$

a. #(2x - 1/2) = (sqr21)/2 -> 2x = 1/2 + (sqr21)/2 ->

$x = 1/4 + (sqr21)/4$

b. $(2x - 1/2) = -(sqr 21)/2 -> 2x = 1/2 - (sqr21)/2$ ->

$x = 1/4 - (sqr21)/4$

How do you solve 4x^2-2x-5=04x^2-2x-5=0 by completing the square?