Recall:

 R 2 = 2 + T T 2 = 2 − V V 2 = 2 − R

Substitute:

R 2 = 2 + √(2 − V)
R 2 = 2 + √(2 − √(2 − R))

For clarity later, change the symbol R to x:

x2 = 2 + √(2 − √(2 − x))

Rearrange, and square both members at the risk of generating extraneous roots:

x2 − 2 = √(2 − √(2 − x))
x4 − 4x2 + 4 = 2 − √(2 − x)

Again, rearrange and square both members:

x4 − 4x2 + 2 = − √(2 − x)
x8 + 16x4 + 4 − 8x6 + 4x4 − 16x2 = 2 − x

Finally, rearrange to form an eighth-degree polynomial:

x8 − 8x6 + 20x4 − 16x2 + x + 2 = 0

which factors into:

(x3x2 − 2x + 1) (x4x3 − 3x2 + 2x + 1) (x + 2) = 0

Three of the roots are pertinent to septrights:

 + 2 cos (180° / 7) = + R − 2 cos (360° / 7) = − T + 2 cos (540° / 7) = + V

But four other roots, which are extraneous, come from not septrights but rather nonarights:

 + 2 cos 20° − 2 cos 40° + 2 cos 60° = + 1 − 2 cos 80°

And one root might go either way:

− 2 cos 0° = − 2

Where did those nonaright values come from? To construct nonarights we can define these:

 q = 2 cos 10° r = 2 cos 20° s = 2 cos 30° t = 2 cos 40° u = 2 cos 50° v = 2 cos 60° w = 2 cos 70° x = 2 cos 80°

which lead to the following formulae, which look very much like the R-T-V formulae above:

 r2 = 2 + t t2 = 2 + x v2 = 2 − v x2 = 2 − r

and that helps explain why they appear as extraneous roots.