How do you factor x^4-7x^2+12=0?

x^4 - 7x^2 + 12

substituting x^2 = t

t^2 - 7t + 12
=> (t-4)(t-3) = 0
=> (x^2 - 4)(x^2-3) = 0

=> (x+2)(x-2)(x+3^0.5)(x-3^0.5) = 0

x^4 - 7x^2 + 12 = 0
x^4 - 3x^2 - 4x^2 + 12 = 0
(x^4 - 3x^2) - (4x^2 - 12) = 0
x^2(x^2 - 3) - 4(x^2 - 3) = 0
(x^2 - 3)(x^2 - 4) = 0
(x^2 - 3)(x + 2)(x - 2) = 0

x^2 - 3 = 0
x^2 = 3
x = ±√3

x + 2 = 0
x = -2

x - 2 = 0
x = 2

∴ x = ±√3 or ±2

( x ² - 4 ) ( x ² - 3 ) = 0

x = ± 2 , x = ±√3

(x^4 - 7x^2 + 12)

= (x^2 - 4)(x^2 - 3)
= (x + 2)(x - 2)(x^2 - 3)

x^4 - 7x^2 + 12 = 0
or, x^4 - 3x^2 - 4x^2 + 12 = 0
or, x^2(x^2 - 3) - 4(x^2 - 3) = 0
or, (x^2 - 3) (x^2 - 4) = 0
or, (x^2 - 3) {(x)^2 - (2)^2}
or, (x^2 - 3) (x + 2) (x - 2) = 0
EITHER, x^2 - 3 =0
=> x^2 = 3
=> x = 3^1/2
OR, x + 2 = 0
=> x = -2
OR, x - 2 = 0
=> x = 2