Problem: Solve the quadratic equation $2x^2 - 3x - 5 = 0$. 1. Level 1 — Identify coefficients and apply the quadratic formula. For $ax^2+bx+c=0$ the solutions are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=2$, $b=-3$, $c=-5$. Compute the discriminant: $\Delta = b^2-4ac = (-3)^2 - 4\cdot 2\cdot (-5) = 9 + 40 = 49$. So $\sqrt{\Delta} = 7$. Plug into the formula: $x = \frac{-(-3) \pm 7}{2\cdot 2} = \frac{3 \pm 7}{4}$. Thus $x = \frac{3+7}{4} = \frac{10}{4} = \frac{5}{2}$ or $x = \frac{3-7}{4} = \frac{-4}{4} = -1$. 2. Level 2 — Verify the solutions by substitution. Substitute $x=\frac{5}{2}$ into $2x^2-3x-5$: $2\left(\frac{5}{2}\right)^2 - 3\left(\frac{5}{2}\right) - 5 = 2\cdot \frac{25}{4} - \frac{15}{2} - 5 = \frac{25}{2} - \frac{15}{2} - \frac{10}{2} = 0$. Substitute $x=-1$: $2\cdot (-1)^2 -3\cdot (-1) -5 = 2 + 3 -5 = 0$. Both satisfy the equation, so they are correct. 3. Level 3 — Graphical interpretation and vertex. The graph $$y=2x^2-3x-5$$ is a parabola opening upward with x-intercepts at $x=-1$ and $x=\frac{5}{2}$. Vertex at $x=\frac{-b}{2a} = \frac{3}{4}$ and $y=2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) -5 = \frac{9}{8} - \frac{9}{4} - 5 = \frac{9}{8} - \frac{18}{8} - \frac{40}{8} = -\frac{49}{8}$. Final answer: $x=-1$ and $x=\frac{5}{2}$.