Infinite Series

Write the expression in a more recognizable form and set it equal to a value $x$ so that $x=\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ . The expression repeats at the second square root, so we can substitute $x$ in that spot. $x=\sqrt{6+x}$ , then square and rearrange to get the quadratic $x^2-x-6=0$ . Solving yields two solutions, $3$ and $-2$ , but you take the positive solution $\boxed{3}$ .

The expression can be written in a more familiar form; $\color{#3D99F6}{(6+(6+(6+(6+(\dotsm)^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}}=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{\dotsm}}}}}$ Now let the above expression be equal to $x$ .Then the expression can be written as; $\color{#20A900}{x=\sqrt{6+x}\\x^2=6+x\\x^2-x-6=0\\ (x-3)(x+2)=0\\x=3\;or\;x=-2}$ As the expression is positive so the answer is $\boxed{3}$

Let:

$x=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6.....}}}}$

$x=\sqrt{6+x}$

Squaring both sides:

$x^{2}=6+x$

$x^{2}-x-6=0$

$x^{2}-3x+2x-6=0$

$x(x-3)+2(x-3)=0$

$(x-3)(x+2)=0$

So, the values of x $=3$ and $-2$

But, $x$ cannot be negative, since square root of negative integers is not possible.

Thus, the answer is: $x=\boxed{3}$