Help finding sin theta using the quadratic formula

Solve the equation in the interval [0degrees, 360degrees)
sin^2theta-6sintheta=4

How do I use the quadratic formula to find the values of sin theta?

I would greatly appreciate any input.

sin²(θ) - 6.sin(θ) = 4

sin²(θ) - 6.sin(θ) - 4 = 0 → let: sin(θ) = x

x² - 6x - 4 = 0

Polynomial like: ax² + bx + c, where:
a = 1
b = - 6
c = - 4

Δ = b² - 4ac (discriminant)

Δ = (- 6)² - 4(1 * - 4) = 36 + 16 = 52 = 4 * 13

x1 = (- b - √Δ) / 2a = (6 - 2√13) / (2 * 1) = 3 - √13

x2 = (- b + √Δ) / 2a = (6 + 2√13) / (2 * 1) = 3 + √13


Recall: sin(θ) = x


First case: sin(θ) = x1

sin(θ) = 3 - √13

sin(θ) ≈ - 0.6055512 → θ ≈ - 37.268522 → θ ≈ 322.73148


Second case: sin(θ) = x2

sin(θ) = 3 + √13 → no possible because: - 1 < sin(θ) < 1

hint: let sin(t) = x


sin(t)^2 - 6sin(t) = 4
x^2 - 6x = 4
x^2 - 6x - 4 = 0
x = (6 +/- sqrt(36 + 16)) / 2
x = (6 +/- sqrt(52)) / 2
x = (6 +/- 2 * sqrt(13)) / 2
x = 3 +/- sqrt(13)
sin(t) = 3 +/- sqrt(13)

3 + sqrt(13) is greater than 1, so there are no real values for t that sill work


sin(t) = 3 - sqrt(13)
t = arcsin(3 - sqrt(13))
t = 322.73147831836018129826819105389 , 217.26852168163981870173180894611

substitute sintheta with x.
then you have x^2-6x-4=0
x= [6 +/- sqrt(36- -16)]/ 2
x= [6 +/- sqrt52] /2
x= 3+sqrt13, x=3-sqrt13
now resubstitute sintheta
sin theta = 3+/- sqrt 13
sintheta= 6.605
sintheta= -0.605 ==> arcsin -0.605= -37.27= 322.73degres

sin ² θ - 6sin θ - 4 = 0

sin θ = [6 ± √ (36 + 16)]/2 = [6 ± √(52)]/2 = 3 ± √(13)

3 + √(13) > 1, so sin θ = 3 - √(13) is the only solution.