Chapter 2 - Linear and Quadratic Functions - Section 2.8 Equations and Inequalities Involving the Absolute Value Function - 2.8 Assess Your Understanding - Page 182: 37

$-2, -1,0,1$.

Step 1. The original equation can be separated into two cases: $x^2+x-1=1$ or $x^2+x-1=-1$. Step 2. For $x^2+x-1=-1$, we have$x^2+x=0 \longrightarrow (x)(x+1)=0 \longrightarrow x=-1,0$. Step 3. For $x^2+x-1=1$, we have $x^2+x-2=0 \longrightarrow (x+2)(x-1)=0 \longrightarrow x=-2,1$.