I prefer to solve quadratic inequalities directly by completing the square, rather than looking at each possible solution set. (I know with higher polynomial and other inequalities this is the only possibility though)...
$\displaystyle \begin{align*} 3x^2 + 16x + 21 &> 0 \\ x^2 + \frac{16}{3}x + 7 &> 0 \\ x^2 + \frac{16}{3}x + \left( \frac{8}{3} \right) ^2 - \left( \frac{8}{3} \right) ^2 + 7 &> 0 \\ \left( x + \frac{8}{3} \right) ^2 - \frac{64}{9} + \frac{63}{9} &> 0 \\ \left( x + \frac{8}{3} \right) ^2 - \frac{1}{9} &> 0 \\ \left( x + \frac{8}{3} \right) ^2 &> \frac{1}{9} \\ \sqrt{ \left( x + \frac{8}{3} \right) ^2 } &> \sqrt{ \frac{1}{9} } \\ \left| x + \frac{8}{3} \right| &> \frac{1}{3} \end{align*}$
I'm sure you can go from here...