Need help finding the solution of sec^2x + 2secx - 8 =0

sec^2x + 2secx - 8 =0
let y = sec x
y² + 2y - 8 =0
use the quadratic formula to solve
y = -2/2 +/- (1/2)(√(4 -4(-8)) = -1 +/- 3 = -4 and 2
sec x = -4 and sec x = 2
sec x = 1/cos x
cos x = -1/4 and cos x = 1/2
x = 104.47° and 60°
and since cos (x) = cos (2π -x)
so 255.53° and 300° are also answers

sec^2 x + 2sec x - 8 = 0

let u = sec x

u^2 + 2u - 8 = 0

u^2 + 2u = 8

complete square

(2/2)^2 = 1^2 = 1 so add 1 to both sides and we get

u^2 + 2u + 1 = 8 + 1

(u + 1)^2 = 9

u + 1 = +- 3

u = - 1 +- 3 or u = - 4 or u = 2 replace u with sec x

sec x = - 4 or sec x = 2

1/cos x = - 4 or 1/cos x = 2

- 1/4 = cos x or 1/2 = cos x

104.47 = x or 60 = x

let sec x = u

u^2+2u-8=0

(u+4)(u-2) = 0

u = -4 and u = 2

so cos x = -1 / 4 and cos x = 1 / 2

x = 60 , 300, 104.48 and 255.22