1. **Problem:** Find the points of intersection of the curve $2x^2 - y + 19 = 0$ and the line $y + 11x = 4$. 2. **Step 1: Express $y$ from the line equation.** From $y + 11x = 4$, we get $$y = 4 - 11x$$ 3. **Step 2: Substitute $y$ into the curve equation.** Substitute $y = 4 - 11x$ into $2x^2 - y + 19 = 0$: $$2x^2 - (4 - 11x) + 19 = 0$$ 4. **Step 3: Simplify the equation.** $$2x^2 - 4 + 11x + 19 = 0$$ $$2x^2 + 11x + 15 = 0$$ 5. **Step 4: Solve the quadratic equation.** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=11$, $c=15$. Calculate the discriminant: $$\Delta = 11^2 - 4 \times 2 \times 15 = 121 - 120 = 1$$ 6. **Step 5: Find the roots.** $$x = \frac{-11 \pm \sqrt{1}}{2 \times 2} = \frac{-11 \pm 1}{4}$$ So, $$x_1 = \frac{-11 + 1}{4} = \frac{-10}{4} = -\frac{5}{2}$$ $$x_2 = \frac{-11 - 1}{4} = \frac{-12}{4} = -3$$ 7. **Step 6: Find corresponding $y$ values.** For $x_1 = -\frac{5}{2}$: $$y = 4 - 11 \times \left(-\frac{5}{2}\right) = 4 + \frac{55}{2} = \frac{8}{2} + \frac{55}{2} = \frac{63}{2}$$ For $x_2 = -3$: $$y = 4 - 11 \times (-3) = 4 + 33 = 37$$ 8. **Final answer:** The points of intersection are $$\left(-\frac{5}{2}, \frac{63}{2}\right) \text{ and } (-3, 37)$$