Will you square ??? Part:2

First we have

$\sqrt { 18+\sqrt { 308 } } =\sqrt { 18+2\sqrt { 77 } } =\sqrt { 11+7+2\sqrt { 11 } \times \sqrt { 7 } } =\sqrt { { \left( \sqrt { 11 } +\sqrt { 7 } \right) }^{ 2 } } =\sqrt { 11 } +\sqrt { 7 }$

Similarly

$\sqrt { 15+\sqrt { 104 } } =\sqrt { 15+2\sqrt { 26 } } =\sqrt { 13+2+2\sqrt { 13 } \times \sqrt { 2 } } =\sqrt { { \left( \sqrt { 13 } +\sqrt { 2 } \right) }^{ 2 } } =\sqrt { 13 } +\sqrt { 2 }$

Therefore

$e\sqrt { a } +f\sqrt { b } +g\sqrt { c } +h\sqrt { d } =\sqrt { 11 } +\sqrt { 7 } +\sqrt { 13 } +\sqrt { 2 }$

Therefore $a+b+c+d=13+2+11+7=33$

First, we need to remove the square root sign to make it simpler to solve.

$\sqrt{18+\sqrt{308}}$

$=\sqrt{18+2\sqrt{77}}$

$=\sqrt{11+7+2\sqrt{77}}$

$=\sqrt{(\sqrt{11})^{2}+(\sqrt{7})^{2}+2\times\sqrt{11}\times\sqrt{7}}$

$=\sqrt{(\sqrt{11}+\sqrt{7})^{2}}=\sqrt{11}+\sqrt{7}$

Similarly, we will solve for the second one.

$\sqrt{15+\sqrt{104}}$

$=\sqrt{15+2\sqrt{26}}$

$=\sqrt{13+2+2\sqrt{26}}$

$=\sqrt{(\sqrt{13})^{2}+(\sqrt{2})^{2}+2\times\sqrt{13}\times\sqrt{2}}$

$=\sqrt{(\sqrt{13}+\sqrt{2})^{2}}=\sqrt{13}+\sqrt{2}$

So, $\sqrt{18+\sqrt{308}}+\sqrt{15+\sqrt{104}}=\sqrt{11}+\sqrt{7}+\sqrt{13}+\sqrt{2}$

Therefore,

$e\sqrt{a}+f\sqrt{b}+g\sqrt{c}+h\sqrt{d}=\sqrt{11}+\sqrt{7}+\sqrt{13}+\sqrt{2}$

Therefore, the answer is: $a+b+c+d=11+7+13+2=\boxed{33}$