2cos2θ-cosθ-1=0 for - 180°≤θ≤180°

I don't get how it is solved

Hello,

cos 2t=2cos²t-1

2cos 2t - cos t-1=

==>2 (2cos² t - 1) - cos t-1=0

==>4cos² t - cos t - 3=0

Δ = (1+4*4*3=49=7²

cos t = (1+7) / 8 or cos t = (1-7) / 8

cos t = 1 or cos t=-3/4

t=0° or t=138.590377 ... °