Answer
$\{-\frac{7}{3}, 2, \frac{-1\pm i\sqrt {167}}{6} \}$
Work Step by Step
Step 1. Given $|3x^2+x|=14$, remove the absolute sign to get $3x^2+x=14$ and $3x^2+x=-14$
Step 2. For $3x^2+x=14$, we have $3x^2+x-14=0$ or $(x-2)(3x+7)=0$ which gives $x=-\frac{7}{3}, 2$
Step 3. For $3x^2+x=-14$, we have $3x^2+x+14=0$ which gives $x=\frac{-1\pm\sqrt {1-4(3)(14)}}{2(3)}=\frac{-1\pm i\sqrt {167}}{6}$
Step 4. Combine the above results, the solution set is $\{-\frac{7}{3}, 2, \frac{-1\pm i\sqrt {167}}{6} \}$
