16x2-8x=1=0?

[05]
16x^2-8x+1=0
(4x)^2-2*4x*1+(1)^2=0
(4x-1)^2=0
4x-1=0
4x=1
x= 1/4

Assuming you meant:

16x^2 - 8x + 1 = 0

factor:

(4x - 1)(4x - 1) = 0

Since both terms are the same we can just set one term equal to 0:

4x - 1 = 0
4x = 1
x = 1/4.

8x 1

if it is 16x² - 8x + 1 = 0 then
16x² - 4x -4x + 1 = 0
4x(4x - 1) -1 ( 4x - 1) = 0
(4x - 1) ( 4x - 1) = 0
(4x - 1)² = 0
4x - 1 = 0
4x = 1
x = 1/4.

Do you mean: 16x^2 - 8x + 1 = 0

16x^2 - 8x + 1 = 0
(4x - 1)(4x - 1) = 0

4x - 1 = 0
4x = 1
x = 1/4

∴ x = 1/4