How much less `2a^(2) + 1` is than `3a^(2) - 6` ?

Answer

To find out how much less \(2a^2 + 1\) is than \(3a^2 - 6\), we need to subtract \(2a^2 + 1\) from \(3a^2 - 6\). Here’s the step-by-step solution:

### Step 1: Write the expression for the difference
We need to find the difference:
\[ (3a^2 - 6) - (2a^2 + 1) \]

### Step 2: Distribute the negative sign
Now, we will distribute the negative sign across the second expression:
\[ 3a^2 - 6 - 2a^2 - 1 \]

### Step 3: Combine like terms
Next, we will combine the like terms:
- Combine \(3a^2\) and \(-2a^2\):
\[ 3a^2 - 2a^2 = 1a^2 \text{ or simply } a^2 \]
- Combine \(-6\) and \(-1\):
\[ -6 - 1 = -7 \]

Putting it all together, we have:
\[ a^2 - 7 \]

### Conclusion
Thus, \(2a^2 + 1\) is \(a^2 - 7\) less than \(3a^2 - 6\).