Find the square of :
`sqrt(3)+ sqrt(2)`

Answer

To find the square of \( \sqrt{3} + \sqrt{2} \), we can follow these steps:

### Step 1: Write the expression
We start with the expression we need to square:
\[ (\sqrt{3} + \sqrt{2})^2 \]

### Step 2: Use the identity for squaring a binomial
We can use the algebraic identity for squaring a binomial, which states:
\[ (a + b)^2 = a^2 + b^2 + 2ab \]
In our case, let \( a = \sqrt{3} \) and \( b = \sqrt{2} \).

### Step 3: Calculate \( a^2 \) and \( b^2 \)
Now we calculate \( a^2 \) and \( b^2 \):
\[ a^2 = (\sqrt{3})^2 = 3 \]
\[ b^2 = (\sqrt{2})^2 = 2 \]

### Step 4: Calculate \( 2ab \)
Next, we calculate \( 2ab \):
\[ 2ab = 2 \times \sqrt{3} \times \sqrt{2} = 2 \times \sqrt{6} \]

### Step 5: Combine all parts
Now we can combine all the parts together:
\[ (\sqrt{3} + \sqrt{2})^2 = a^2 + b^2 + 2ab = 3 + 2 + 2\sqrt{6} \]
\[ = 5 + 2\sqrt{6} \]

### Final Answer
Thus, the square of \( \sqrt{3} + \sqrt{2} \) is:
\[ \boxed{5 + 2\sqrt{6}} \]