Solution
= Solution
\b[True.] For every <integer> $n\geq1$, $n^{3/2}\leq n^2$ and $1\leq n^2$, so
$$
0\leq3n^2+5n^{3/2}+6\leq(3+5+6)n^2=14n^2.
$$
Thus $C=14$ and $N=1$ witness the required <Big O notation> bound.
= Solution
\b[True.] For every <integer> $n\geq1$, $n^{3/2}\leq n^2$ and $1\leq n^2$, so
$$
0\leq3n^2+5n^{3/2}+6\leq(3+5+6)n^2=14n^2.
$$
Thus $C=14$ and $N=1$ witness the required <Big O notation> bound.