Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Practice - Page 456: 8

The solutions are $-3$ , $-\frac{1}{3}$ and $7$

We simplify as follows: $(x+3)(3x^{2}-20x-7)=0$ $(x+3)(3x+1)(x-7)=0$ $x+3=0$ or $3x+1=0$ or $x-7=0$ $x=-3$ or $3x=-1$ or $x=7$ $x=-3$ or $x=-\frac{1}{3}$ or $x=7$