solve this equation as an reducible to the quadratic

(2x-4)^2-4(2x-4)-5=0
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2 Answers

let 2x-4=y & yu hav y^2 -4y -5=0

quadratik equashun giv (y-5)*(y+1)=0

thus, 2x-4=5 or -1

2x-4=5...2x=9, x=4.5

2x-4=-1...2x=3, x=1.5
by

Given (2x-4)^2-4(2x-4)-5=0
4x^2+72=24x
4x^2-24x+27=0
(2x-9)(2x-3)=0
2x-9=0 or 2x-3=0
x=9/2 or x=3/2

Elementry Algbera - http://math.tutorcircle.com/algebra/#elementary-algebra

by Level 8 User (30.1k points)

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