Base case: For n = 7, you have

3 n^{2}= 3 * 49 = 147

17n + 10 = 17 * 7 + 10 = 129

So, 3n^{2}>17n+10

Induction: Assume the inequality holds for n = k > 7. That is, assume that 3k^{2}>17k+10. We need to prove that the inequality holds for n = k +1. That is, we need to prove that 3(k+1)^{2}>17(k+1)+10.

But 3(k+1)^{2}= 3k^{2}+ 6k + 3. We know from our assumption 3k^{2}>17k+10. So, we get 3(k+1)^{2}= 3k^{2}+ 6k + 3 > 17k + 10 + 6k + 3 > 17k + 17 + 10 = 17(k+1) + 10. (Note: 6k+3 > 17 when k > 7.)