How do you simplify $6\sqrt{7}+2\sqrt{28}$?

Hint: We try to simplify the surd value and convert to the same form so that the addition can be done easily. The surd form of $\sqrt{7}$ is already in simplified form. We need to convert $\sqrt{28}$. The addition happens for the coefficients.

Complete step by step solution:
The given addition is in the form of surds. We first need to simplify the complex surds into their simplified form.
The surd form of $\sqrt{7}$ is already in simplified form.
We just need to find the form of $\sqrt{28}$.
We need to find the prime factorisation of the given number 28.
$\begin{align}
  & 2\left| \!{\underline {\,
  28 \,}} \right. \\
 & 2\left| \!{\underline {\,
  14 \,}} \right. \\
 & 1\left| \!{\underline {\,
  7 \,}} \right. \\
\end{align}$
Therefore, \[28=2\times 2\times 7\].
For finding the square root, we need to take one digit out of the two same number of primes.
So, the multiplication is \[\sqrt{28}=\sqrt{2\times 2\times 7}=2\sqrt{7}\].
We can write $2\sqrt{28}=4\sqrt{7}$. Therefore, $6\sqrt{7}+2\sqrt{28}=6\sqrt{7}+4\sqrt{7}$
Now for the addition the surd value for both of the terms is the same.
The coefficients are 6 and 4 and both of them are positive.
We add them to find the final coefficient value. So, $6+4=10$.
$6\sqrt{7}+2\sqrt{28}=6\sqrt{7}+4\sqrt{7}=10\sqrt{7}$.
Therefore, the simplified form of $6\sqrt{7}+2\sqrt{28}$ is $10\sqrt{7}$.

Note: We need to remember that the addition of simplified unequal surds is not possible. For example, we can’t simplify the addition $\sqrt{7}+\sqrt{5}$. They are already in their simplified form and they are unequal.