#### Answer

$\left\{-\dfrac{7}{2}, -\dfrac{4}{3}\right\}$

#### Work Step by Step

Solve the equation using the $\text{ac-method}$.
With $a=6$ and $c=28$, $ac= 6(28)=168$
Recall:
$ax^2+bx+c$ can be factored if there exists $d$ and $e$ such that $ac=de$ and $d+e=b$.
The trinomial can be factored by grouping by rewriting the trinomial as $ax^2+dx+ex+c$.
Look for factors of $168$ whose sum is $29$
Note that $168=21(8)$ and $21+8=29$.
This means that the trinomial is factorable.
Solve the given equation by grouping:
\begin{align*}
6w^2+29w+28&=0\\
6w^2+21w+8w+28&=0\\
(6w^2+21w)+(8w+28)&=0\\
3w(2w+7)+4(2w+7)&=0\\
(2w+7)(3w+4)&=0\end{align*}
Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain:
\begin{align*}
2w+7&=0 &\text{or}& &3w+4=0\\
2w&=-7 &\text{or}& &3w=-4\\
w&=-\frac{7}{2} &\text{or}& &3w=-\frac{4}{3}\\
\end{align*}