Algebra

The following is using brute force because I sux at math. ;)

First, let's combine some like terms within the parenthesis. We have

$\left(2\sqrt{3}+\sqrt{5}+7\right)^{2}$

Since we can't simplify this forwards, let's set up the squaring process. Expanding the square, we get

$\left(2\sqrt{3}+\sqrt{5}+7\right)\left(2\sqrt{3}+\sqrt{5}+7\right)$

Now, let's use distributive property. Distributing every number in, and simplifying ALOT, we get

${2\sqrt{3}\cdot \left(2\sqrt{3}+\sqrt{5}+7\right)+\sqrt{5}\cdot \left(2\sqrt{3}+\sqrt{5}+7\right)+7\left(2\sqrt{3}+\sqrt{5}+7\right)}\\$

I'm not going to show the entire process, but I think you got it!

Now, remember that a square being square becomes an integer. 

Distributing every single number in and doing some simplifying, we get

$2\sqrt{15}+14\sqrt{3}+12+2\sqrt{15}+7\sqrt{5}+5+{14\sqrt{3}+7\sqrt{5}+49}$

$4\sqrt{15}+28\sqrt{3}+14\sqrt{5}+66$

This is the furthest we can simplify, so it's our final answer. 

Thanks! :)