Solve for x: x=√(2+√(2+√(2...))) See the picture on the details for a good view of the equation.

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and this

Solve for Y: √(y+√(y+√(y...))) = 5

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x=√(2+√(2+√(2...)))
Hence
x=√(2+ [√(2+√(2...))] )
x=√(2+ x)
x≥0 and x≥-2 => x≥0
x²=2+ x => x=2


√(y+√(y+√(y...))) = 5
√(y+[ √(y+√(y...))] ) = 5
√(y+5) = 5
y≥-5
y+5 = 25 => y=20

x = √(2+√(2+√(2...)))
x² = 2 +√(2+√(2...)))
x² = 2 + x
x² – x – 2 = 0
(x – 2)(x + 1) = 0
x = 2 or – 1

√(y+√(y+√(y...))) = 5
y² + y = 5
y² + y – 5 = 0
y = [– 1 ± √1 + 5)]/2
= [– 1 ± √1 + 5)]/2
= [– 1 ± √6)]/2
= –1/2 ± √6/2
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