Given that $\displaystyle \sum^n_{r=1}r(2r-1)=\frac{1}{6}n(n+1)(4n-1)$

Find the set of values of $\displaystyle n$ for which $\displaystyle \sum^n_{r=1}(6r^2-3r+2^r)>10000$

$\displaystyle \sum^n_{r=1}(6r^2-3r+2^r)=3\sum^n_{r=1}r(2r-1)+\sum^n_{r=1}2^r$

$\displaystyle =3[\frac{1}{6}n(n+1)(4n-1)]+\frac{2(1-2^n)}{1-2}$

$\displaystyle =\frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}$

How do I solve for n?