To solve the expression $6\sqrt{3} \times 2\sqrt{15}$, follow these steps:

1. Multiply the constants: Multiply the numbers outside the square roots. $6 \times 2 = 12$

2. Multiply the square roots: Multiply the numbers inside the square roots. $\sqrt{3} \times \sqrt{15} = \sqrt{3 \times 15} = \sqrt{45}$

3. Simplify the square root: Simplify $\sqrt{45}$.

   • Factor $45$ as $9 \times 5$.
   • So, $\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$.

4. Combine the results: Multiply $12$ by $3\sqrt{5}$. $12 \times 3\sqrt{5} = 36\sqrt{5}$

Therefore, the final answer is:

$\boxed{36\sqrt{5}}$

Would you like further details or have any questions?

Here are some related questions:

1. How do you multiply square roots with different radicands?
2. What are the steps to simplify square roots in multiplication?
3. Can you explain how to factor numbers to simplify square roots?
4. How do you handle negative numbers under a square root in multiplication?
5. What is the significance of simplifying square roots in algebra?

Tip: Always factor numbers inside the square root into perfect squares when possible to simplify expressions easily!