Answer
The solution set is $\left\{-\frac{5}{2}, 0, \frac{3}{7} \right\}$.
Work Step by Step
Simplify the equation to obtain:
\begin{align*}
14x^3+35x^2-20x+10-6x^2+5x-10&=0\\
14x^3+29x^2-15x&=0
\end{align*}
Factor out $x$:
$$x(14x^2+29x-15)=0$$
Factor the trinomial to obtain:
$$x(7x-3)(2x+5)=0$$
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
x&=0 &\text{or}& &7x-3=0& &\text{or}& &2x+5=0\\
x&=0 &\text{or}& &7x=3& &\text{or}& &2x=-5\\
x&=0 &\text{or}& &x=\frac{3}{7}& &\text{or}& &x=-\frac{5}{2}\\
\end{align*}
Thus, the solution set is $\left\{-\frac{5}{2}, 0, \frac{3}{7} \right\}$.
