Question 1199930
sqrt[12+sqrt(12+sqrt(12+...] can be shown as x = sqrt(12+x)

When both sides are squared, the equation becomes x^2 = 12 + x

x^2 = 12 + x
0 = x^2 + x + 12
0 = (x+3)(x-4)

In order for the equation to hold true, x must either be -3, or 4

If we substitute -3 and 4 into x = sqrt(12+x), we find that -3 does not satisfy our equation while 4 does. 

Therefore sqrt[12+sqrt(12+sqrt(12+...] is equal to 4