1. **State the problem:** Solve the quadratic equations using the quadratic formula and leave answers in simplified radical form. 2. **Recall the quadratic formula:** For an equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients. 3. **Solve the first equation:** $x^2 + 3x - 11 = 0$ - Here, $a=1$, $b=3$, $c=-11$ - Calculate the discriminant: $$b^2 - 4ac = 3^2 - 4(1)(-11) = 9 + 44 = 53$$ - Apply the quadratic formula: $$x = \frac{-3 \pm \sqrt{53}}{2 \times 1} = \frac{-3 \pm \sqrt{53}}{2}$$ 4. **Solve the second equation:** $x^2 - 3x = 8$ - Rewrite as $x^2 - 3x - 8 = 0$ - Here, $a=1$, $b=-3$, $c=-8$ - Calculate the discriminant: $$b^2 - 4ac = (-3)^2 - 4(1)(-8) = 9 + 32 = 41$$ - Apply the quadratic formula: $$x = \frac{-(-3) \pm \sqrt{41}}{2 \times 1} = \frac{3 \pm \sqrt{41}}{2}$$ **Final answers:** - For $x^2 + 3x - 11 = 0$, $$x = \frac{-3 \pm \sqrt{53}}{2}$$ - For $x^2 - 3x = 8$, $$x = \frac{3 \pm \sqrt{41}}{2}$$