1. Problem statement. Problem: Solve the inequality $2x^2+5x+2>0$. 2. Formula and rules. For a quadratic $ax^2+bx+c>0$ with $a>0$ the parabola opens upward and the quadratic is positive for $x$ outside the real roots. 3. Compute discriminant and roots using the quadratic formula. Compute discriminant: $\Delta=b^2-4ac=5^2-4\cdot2\cdot2=25-16=9$. The roots are $x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-5\pm3}{4}$. Simplify the first root: $\frac{-5+3}{4}=\frac{-2}{4}$. $$\frac{-2}{4}=\frac{\cancel{2}(-1)}{\cancel{2}2}=-\frac{1}{2}$$ Simplify the second root: $\frac{-5-3}{4}=\frac{-8}{4}$. $$\frac{-8}{4}=\frac{\cancel{4}(-2)}{\cancel{4}1}=-2$$ Therefore the roots are $x=-\frac{1}{2}$ and $x=-2$. 4. Factorization check. We can factor the quadratic as $$2x^2+5x+2=(2x+1)(x+2)$$ This matches the computed roots. 5. Sign analysis and solution. Since $a=2>0$ the parabola opens upward, so the quadratic is positive for $x$ less than the smaller root and greater than the larger root. Hence the solution set is $$(-\infty,-2)\cup\left(-\frac{1}{2},\infty\right)$$. Final answer: $(-\infty,-2)\cup\left(-\frac{1}{2},\infty\right)$.