how to solve this? 3(x-5)^2 - 12 = 0?

3 (x - 5)² = 12
(x - 5)² = 4
x - 5 = ± 2
x = 7 , x = 3

3(x-5)² - 12 = 0
=> 3(x² + 5² - 2*5*x) - 12 = 0 [because (a-b)² = a² + b² - 2ab]
=> 3[(x² + 5² - 2*5*x) - 4] = 0 [taking 3 common from the whole equation]
=> (x² + 5² - 2*5*x) - 4 = 0/3 = 0
=> x² - 10x + 25 - 4 = 0
=> x² - 10x + 21 = 0
=> x² - 3x - 7x + 21 = 0 [splitting the middle term]
=> x(x-3) - 7(x-3) = 0
=> (x-3) (x-7) = 0

x-3 = 0
=> x = 3
OR
x-7 = 0
=> x = 7

so the two values of x, on solving the quadratic equation are x = 7 and x = 3

3(x - 5)^2 - 12 = 0
3(x - 5)^2 = 12
(x - 5)^2 = 12/3
(x - 5)^2 = 4
x - 5 = ±√4
x - 5 = ±2

x - 5 = 2
x = 2 + 5
x = 7

x - 5 = -2
x = -2 + 5
x = 3

∴ x = 3 , 7

3(x^2-10x+25) - 12 = 0
3x^2 - 30x + 75 - 12 = 0
3x^2 -30x + 63 = 0
x^2 - 10x + 21 = 0
(x - 7)(x - 3) = 0
x - 7 = 0 or x - 3 = 0
x = 7 or x = 3