已知：正数数列{an}前n项和为Sn,满足6Sn=（an）^2+3an-10,求通项an （n是下标）

n=1时6a1=（a1）^2+3a1-10 a1=5(负项舍去)6Sn=（an）^2+3an-106Sn-1=（an-1）^2+3an-1-10两式相减6an=（an）^2+3an-（an-1）^2-3an-1(an+an-1)(an-an-1)-3(an+an-1)=0an+an-1=0或者an-an-1-3=0由于是正项数列所以an+...