What is the nth term of quadratic sequence 4 7 12 19 28

We find the first differences between terms:
7-4=3; 12-7=5; 19-12=7; 28-19=9.
Since these are different, this is not linear.
We now find the second differences:
5-3=2; 7-5=2; 9-7=2. Then:

Since these are the same, this sequence is quadratic.
We use (1/2a)n², where a is the second difference:
(1/2*2)n²=1n².

We now use the term number of each term for n:
4 is the 1st term; 1*1²=1.
7 is the 2nd term; 1*2²=4.
12 is the 3rd term; 1*3²=9.
19 is the 4th term; 1*4²=16.
28 is the 5th term: 1*5²=25.

Now we find the difference between the actual terms of the sequence and the numbers we just found:

4-1=3; 7-4=3; 12-9=3; 19-16=3; 28-25=3.

Since this is constant, the sequence is in the form (1/2a)n²+d;
in our case, 1n²+d, and since d=3, 1n²+3.
The correct answer is n²+3