Verified answer

Assuming that you mean √(132 + √(132 + √(132 + ...))):

Let x = √(132 + √(132 + √(132 + ...))).

Then, we have

x = √(132 + √(132 + √(132 + ...)))

...= √(132 + x).

Squaring both sides gives us a polynomial equation:

x^2 = 132 + x

==> x^2 - x - 132 = 0

==> (x - 12)(x + 11) = 0

==> x = 12, since x = √(132 + √(132 + √(132 + ...))) > 0.

I hope this helps!

assuming this is a nested radical:

a = √(132+√(132+√(132+..

a = √(132+a

a² = 132+a

a²-a = 132

k=132

a²-a = k

a²-a+1/4 = k+1/4

(a-1/2)² = (4k+1)/4

a= 1/2 ±√(4k+1)/2

a=(1±√(4k+1))/2

4k+1=529=23²

a=(1±23)/2

a=24/2,-22/2=12,-11 (reject < 0)

a²-a-132=0

132=2²•3•11

(a-12)(a+11)=0

a=12,-11 (reject < 0)

answer 12

Thanks for the 2 points loser