Chapter 3 - Linear and Quadratic Functions - 3.5 Inequalities Involving Quadratic Functions - 3.5 Assess Your Understanding - Page 161: 41

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$x^2+x+1>0$, since its determinant ($b^2-4ac=(1)^2-4\cdot1\cdot1=-3$) is negative ($-3$), it has only positive values (also because the leading coefficient, $1$ is positive), hence the solution set is all real numbers. (also, see the graph of $x^2+x+1$)