Heyy! In scenarios like this, the \(a\) coefficient isn't \(1\), so I would either factor it using decomposition or use the quadratic formula.
here's a article on decomposition (https://tentotwelvemath.com/fmp-10/factors-and-products/factoring-a-trinomial-using-decomposition/)
here's a article on the quadratic formula (https://www.purplemath.com/modules/quadform.htm)
Hoped that helped!
Here is a massive hint!
Any 2-digit number with a 3 is peachy (and all peachy numbers have at least one 3). There are nine 2-digit numbers with ones digit 3, and there are nine additional 2-digit numbers with tens digit 3 (except from 33, which we've already counted, since its ones digits is 3). Therefore, there are . . .