To find the first four terms of the sequence defined by the formula $a_n = 4n^2 + 3n - 6$, we will substitute $n = 1, 2, 3, 4$ into the given formula.

1. For $n = 1$: $a_1 = 4(1)^2 + 3(1) - 6 = 4 + 3 - 6 = 1$ Thus, the first term $a_1 = 1$.

2. For $n = 2$: $a_2 = 4(2)^2 + 3(2) - 6 = 4(4) + 6 - 6 = 16 + 6 - 6 = 16$ Thus, the second term $a_2 = 16$.

3. For $n = 3$: $a_3 = 4(3)^2 + 3(3) - 6 = 4(9) + 9 - 6 = 36 + 9 - 6 = 39$ Thus, the third term $a_3 = 39$.

4. For $n = 4$: $a_4 = 4(4)^2 + 3(4) - 6 = 4(16) + 12 - 6 = 64 + 12 - 6 = 70$ Thus, the fourth term $a_4 = 70$.

Conclusion:

The first four terms of the sequence are: $a_1 = 1, \ a_2 = 16, \ a_3 = 39, \ a_4 = 70.$

Would you like further details on how this was calculated, or do you have any questions?

Here are some related questions to consider:

1. What is the 5th term of this sequence?
2. How would you describe the growth rate of this sequence?
3. What is the difference between consecutive terms in this sequence?
4. Can the formula be written in a different form to identify patterns more easily?
5. What is the general behavior of quadratic sequences like this one?

Tip: When working with quadratic sequences, the second differences between consecutive terms will always be constant.