x^2 - 16x + 48 = 0?

(x² - 16x + 64) = - 48 + 64
(x - 8)² = 16
(x - 8) = ± 4
x = 12 , x = 4
OR
(x - 4)(x - 12) = 0
x = 4 , x = 12

"completeing the square" means that you take an equation in standard form: Ax² + Bx + C = 0 and manipulate "Ax² + Bx + C" so that when you factor it, you get
(x ± α)(x ± α)
(x ± α)² <-- this is what makes it a perfect square equation... the factors are equal.



for Ax² + Bx + C = 0 to be a perfect square, C must = (B/2)²
if (B/2)²≠C simply move C to the other side of the = and then add (B/2)² to both sides, this is your "new C"

so take half of 16 and square that:
16/2 = 8
8² = 64

64≠48 so:
x²-16x = -48 <-- move original C
x² - 16x + 64 = -48 + 64 <--add "new C" to both sides (64)
x² - 16x + 64 = 16
(x-8)(x-8) = 16
(x-8)² = 16
x-8 = ±√16
x-8 = ±4
x = 8±4
x = 4, 12

x^2-16x+(16/2)^2-(16/2)^2+48=0
x^2-16x+(16/2)^2=8^2-48
(x-8)^2=16
(x-8)=+/-4

x-8=4, x=12
x-8=-4, x=4

x^(2)-16x+48=0
=>x^(2)-16x+48-64+64=0
=>(x^(2)-16x+64)-64+48=0
=>(x-8)^2-16=0
=>(x-8)^2=16
=>(x-8)=(+)or(-) 4
=>x=12 ,4

x^2 - 16x + 48 =0
x^2 - 12x - 4x + 48 = 0
x(x - 12) - 4(x - 12) = 0
(x - 4) (x - 12) = 0
either x - 4 = 0 or x - 12 = 0
x = 4 or x = 12

X^2-16X+48=0
X^2-2*8X+8^2-16=0
(x-8)^2-4^2=0
(x-8-4)(x-8+4)=0
(x-12)(x-4)=0
x=12
x=4

x^2 - 16x + 48 = 0
x^2 - (12x -4x) + 48 = 0
[x (x-12)] - [4 (x-12)] = 0
(x - 4)(x - 12) = 0

Either
x - 4 = 0
x = 4

Or

x - 12 = 0
x = 12

x = 4 or 12

x^2 - 16x + 48 = 0
x^2 - 12x -4x+ 48 = 0
x(x-12)-4(x-12) = 0
(x-12)(x-4)=0
so
x-12=0
x=12
or
x-4=0
x=4

so x=4,12

x^2 - 16x + 48 = 0
(x - 4)(x - 12) = 0

x - 4 = 0
x = 4

x - 12 = 0
x = 12

∴ x = 4 , 12