Question 60418
x=sqrt(4+sqrt(4-sqrt(4+sqrt(4-...
{{{x=sqrt(4+sqrt(4-x)) }}}
Then, {{{x^2=4+sqrt(4-x)}}}
Or, {{{x^2-4=sqrt(4-x)}}}
Or, {{{(x^2-4)^2=4-x}}}
Or, {{{(x^2-4)^2+x-4=0}}}
Or, {{{x^4-8x^2+x+12 = 0}}}

This is a 4th degree, or quartic, polynomial.

There's a quartic equation solver here: http://www.1728.com/quartic.htm
 
Plug in 1, 0, -8, 1, and 12 as values for A, B, C, D, E. The value for X2 (2.3027756477...) is correct in that that value of x satisfies the equation: {{{x=sqrt(4+sqrt(4-x))}}}  and the sum looks like it's a bit bigger then 2.