# Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 362: 85

The solution set is $\left\{-\frac{7}{2}, 0, 1\right\}$.

#### Work Step by Step

Factor out $n$: $$n(2n^2+5n-7)=0$$ Factor the trinomial to obtain: $$n(2n+7)(n-1)=0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} n&=0 &\text{or}& &2n+7=0& &\text{or}& &n-1=0\\ n&=0 &\text{or}& &2n=-7& &\text{or}& &n=1\\ n&=0 &\text{or}& &n=-\frac{7}{2}& &\text{or}& &n=1\\ \end{align*} Thus, the solution set is $\left\{-\frac{7}{2}, 0, 1\right\}$.

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