How to solve 4x^2 -24x + 1 = 0 by completing the square

4x^2 - 24x + 1 = 0

divide by 4

x ^2 - 6x = -1/4

add 9 both sides

x^2 - 6x + 9 = 35/4

=> (x - 3)^2 = 35/4

x - 3 = ± 1/2√35

x = 3 ± 1/2√35

4x² - 24x + 1 = 0         Step one: make the "x²" coeficient +1
x² - 6x + ¼ = 0            Take ½ the "x" coefficient, square it,
x² - 6x + 9 = 9 - ¼       then add to both sides of "=." Move the ¼
(x - 3)² = 9 - ¼            "Square-root" both sides of "="
x - 3 = ±√(9 - ¼)          Move the "3" over. You can stop here,
x = 3 ±√(9 - ¼)            or clean things up a bit. Yeah! You're done!

x = 3 ± ½√35

4x^2 - 24x + 1 = 0
or, (2x)^2 -2*2x*6 +36 -35 =0
or, (2x-6)^2 -35=0
or, (2x-6)^2=35
or, 2x-6=+-sqrt(35)
either 2x-6=+sqrt(35) or, 2x-6= - sqrt(35)
2x=6+sqrt(35) 2x=6 - sqrt(35)
ie, x=0.5[6+sqrt(35)] x=0.5[6 - sqrt(35)]

4x^2 - 24x + 1 = 0
(2x)^2 - 2 ( 2 × 6x) + 1 = 0
(2x)^2 - 2 ( 2 × 6x) + 36 + 1 = 0
4x^2 - 24x + 36 – 36 + 1 = 0
(4x^2 - 24x + 36) – 35 = 0
(2x – 6)² = 35
2x – 6 = ± √35
x = 6/2 ± (√35)/2
x = 3 ± (√35)/2
= 3 ± 5.92
x = 8.92 or – 2.92
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4x^2 -24x + 1 = 0
4( x^2 -6x +1/4 )=0
x^2 -6x +1/4 = 0
(x -3)^2 - 9 +1/4 =0
(x -3)^2 = 35/4
x -3 = (+ or - ) sqr (35) / 2
x = 3 (+ or - ) sqr (35) / 2

Hope it will be useful :)

4x^2 - 24x + 1 = 0
x^2 - 6x + 1/4 = 0
x^2 - 6x + 9 + 1/4 = 9
(x - 3)^2 + 1/4 = 9
(x - 3)^2 = 35/4
x - 3 = √35/2 or x - 3 = -√35/2
x = 3 + √35/2 or x = 3 - √35/2