Infinite Nested Radical

Let

$S=\sqrt{6+\sqrt{3+\sqrt{\frac{3}{2}+...}}}$

Now note that $\frac{S}{\sqrt{2}}= \sqrt{3+\sqrt{\frac{3}{2}+...}}$ $\Rightarrow S=\sqrt{6+\frac{S}{\sqrt{2}}}$

Square and solve the quadratic in S to get S= $\boxed{2.828}$

first $\sqrt{\frac {3}{2}+\sqrt{\frac {3}{8}+\sqrt{\frac {3}{128}+\sqrt{..... \infty}}}}$ can be factorised to $\sqrt{\frac {3}{2}+\frac{1}{2}\sqrt{\frac {3}{2}+\frac{1}{2}\sqrt{\frac {3}{2}+\frac{1}{2}....\infty}}}$ we get that $\sqrt{\frac {3}{2}+\frac{1}{2}\sqrt{\frac {3}{2}+\frac{1}{2}\sqrt{\frac {3}{2}+\frac{1}{2}....\infty}}} = 1.5$ now substiute $\sqrt {6+\sqrt{3+\sqrt{\frac {3}{2}+\sqrt{\frac {3}{8}+\sqrt{\frac {3}{128}+\sqrt{..... \infty}}}}}} = \sqrt {6+\sqrt{3+1.5}}$ now solve by calculator $\sqrt{6+\sqrt{4.5}}= \sqrt{6+\frac{3\sqrt{2}}{2}}=\sqrt{\frac{12+3\sqrt{2}}{2}}= \boxed{2.849}$

Let $p=\sqrt{6+\sqrt{3+\sqrt{3/2+\sqrt{3/128+...}}}}$ ,

so $p²=6+\sqrt{3+\sqrt{3/2+\sqrt{3/8+\sqrt{3/128+...}}}}$

From that ${p²-6}\sqrt2=\sqrt{6+\sqrt{6+\sqrt{6+...}}}=3$

.So we get $p²=6+3/\sqrt2={12+3\sqrt2}/2$

so $p=\sqrt{{12+3\sqrt2}/2}$

and p is approximately 2.849...