1. **State the problem:** We need to find the solutions to the equation $$-\frac{1}{2}x^2 - x + 8 = x + 3$$ which means finding the values of $x$ where the quadratic function and the line intersect. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$-\frac{1}{2}x^2 - x + 8 - x - 3 = 0$$ 3. **Simplify the equation:** Combine like terms: $$-\frac{1}{2}x^2 - 2x + 5 = 0$$ 4. **Multiply through by -2 to clear the fraction:** $$x^2 + 4x - 10 = 0$$ 5. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=4$, $c=-10$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 4^2 - 4 \times 1 \times (-10) = 16 + 40 = 56$$ 7. **Find the roots:** $$x = \frac{-4 \pm \sqrt{56}}{2} = \frac{-4 \pm 2\sqrt{14}}{2} = -2 \pm \sqrt{14}$$ 8. **Approximate the solutions:** $$\sqrt{14} \approx 3.74$$ So, $$x_1 = -2 + 3.74 = 1.74$$ $$x_2 = -2 - 3.74 = -5.74$$ **Final answer:** The solutions to the equation are approximately $$x \approx 1.74$$ and $$x \approx -5.74$$.