Question

solve the equation as an equation reducible to the equation. (2x-4)^2-4(2x-4)-5=0

Answer 1

so 2.5?

Answer 2

(2x-4)^2 - 4(2x-4) - 5 = 0
Let t = (2x-4)
t^2 - 4t - 5 = 0
Solve the quadratic for t.
Then replace t with 2x-4 and solve for x.

Answer 3

dont quite get it. could you brake it down for me.

Answer 4

(2x-4)^2 - 4(2x-4) - 5 = 0
Let t = (2x-4). Replace (2x-4) by t in the above equation:
t^2 - 4t - 5 = 0
Solve for t. First factor the above quadratic equation:
t^2 - 4t - 5 = (t - 5)(t +1) = 0
t = 5 or t = -1
put back 2x - 4 in the place of t:
2x - 4 = 5 or 2x - 4 = -1
2x = 9 or 2x = 3
x = 9/2 or x = 3/2