1. Problem statement: Find the solution set of the inequality $2x^2+5x+2>0$. 2. Formula and rule: To solve a quadratic inequality $ax^2+bx+c>0$ we find its roots and determine the sign of the quadratic on intervals determined by the roots. 3. Factorization: The quadratic factors as $$2x^2+5x+2=(2x+1)(x+2)$$. 4. Roots: Solve $(2x+1)(x+2)=0$. 5. Root calculations: From $2x+1=0$ we get $x=-\tfrac{1}{2}$. 6. Root calculations: From $x+2=0$ we get $x=-2$. 7. Sign analysis: The roots split the real line into three intervals $(-\infty,-2)$, $(-2,-\tfrac{1}{2})$, and $(-\tfrac{1}{2},\infty)$. 8. Test point in $(-\infty,-2)$: Take $x=-3$ then $(2x+1)(x+2)=(2(-3)+1)((-3)+2)=-5\cdot -1=5>0$ so the quadratic is positive there. 9. Test point in $(-2,-\tfrac{1}{2})$: Take $x=-1$ then $(2x+1)(x+2)=(2(-1)+1)((-1)+2)=-1\cdot 1=-1<0$ so negative there. 10. Test point in $(-\tfrac{1}{2},\infty)$: Take $x=0$ then $(2x+1)(x+2)=1\cdot 2=2>0$ so positive there. 11. Conclusion: The inequality $2x^2+5x+2>0$ holds for $x\in(-\infty,-2)\cup(-\tfrac{1}{2},\infty)$. 12. Final answer: $\{x\mid x<-2\text{ or }x>-\tfrac{1}{2}\}$.