Let

29+12\sqrt{5} = (a+b\sqrt{5})^2

29-12\sqrt{5} = (a-b\sqrt{5})^2

in which a and b are positive integers

Addition of the two equations gives

(a+b\sqrt{5})^2+(a-b\sqrt{5})^2 = 58

Expand and simplify the equation

2a^2+2\cdot 5b^2=58

Multiplying the two equations gives

(29+12\sqrt{5})(29-12\sqrt{5}) = (a+b\sqrt{5})^2 (a-b\sqrt{5})^2

Simplify the equation using difference of square formula

(a^2 - b^2\cdot 5)^2 = 29^2 - 12^2\cdot 5=121

Taking the square root

2 systems of equations are obtained from (1) and (2)

Case 1

\begin{cases} a^2+5b^2&=29 \\ a^2 - 5b^2 &=11 \end{cases}

Solving the system of equaitons gives

a = \sqrt{\dfrac{29+11}{2}} =\sqrt{20} = 2\sqrt{5}

b = \sqrt{\dfrac{29-11}{2\cdot 5}}=\sqrt{\dfrac{9}{5}}=3\sqrt{\dfrac{1}{5}}

Case 2

\begin{cases} a^2+5b^2&=29\\ a^2 - 5b^2 &= -11 \end{cases}

Solving the system of equaitons gives

a = 3, b=2

Both results give the same root. Therefore,

The value of the square root of

\sqrt{29+12\sqrt{5}}=\sqrt{(3+2\sqrt{5})^2}=3+2\sqrt{5}

\sqrt{29-12\sqrt{5}}=\sqrt{(3-2\sqrt{5})^2}=2\sqrt{5}-3