Let $$x = 3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$$

So, $$x = 3 + {1 \over {4 + {1 \over x}}} = 3 + {1 \over {{{4x + 1} \over x}}}$$

$$ \Rightarrow (x - 3) = {x \over {(4x + 1)}}$$

$$ \Rightarrow (4x + 1)(x - 3) = x$$

$$ \Rightarrow 4{x^2} - 12x + x - 3 = x$$

$$ \Rightarrow 4{x^2} - 12x - 3 = 0$$

$$x = {{12 \pm \sqrt {{{(12)}^2} + 12 \times 4} } \over {2 \times 4}} = {{12 \pm \sqrt {12(16)} } \over 8}$$

$$ = {{12 \pm 4 \times 2\sqrt 3 } \over 8} = {{3 \pm 2\sqrt 3 } \over 2}$$

$$x = {3 \over 2} \pm \sqrt 3 = 1.5 \pm \sqrt 3 $$.

But only positive value is accepted

So, x = $$1.5 + \sqrt 3 $$