1. **Problem 1: Solve the quadratic equation** $x^2 + 8x + 15 = 0$. 2. The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Here, $a=1$, $b=8$, and $c=15$. 3. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = 8^2 - 4 \times 1 \times 15 = 64 - 60 = 4$$ 5. Since $\Delta > 0$, there are two real roots. 6. Calculate the roots: $$x = \frac{-8 \pm \sqrt{4}}{2 \times 1} = \frac{-8 \pm 2}{2}$$ 7. Roots are: $$x_1 = \frac{-8 + 2}{2} = \frac{-6}{2} = -3$$ $$x_2 = \frac{-8 - 2}{2} = \frac{-10}{2} = -5$$ --- 1. **Problem 2: Simplify** $2\omega + 2\omega^2$ 2. Factor out the common factor 2: $$2\omega + 2\omega^2 = 2(\omega + \omega^2)$$ 3. This is the simplified form unless more information about $\omega$ is given. --- 1. **Problem 3: Calculate** $\cos 60^\circ$ 2. Recall that $\cos 60^\circ = \frac{1}{2}$. 3. So, the value is $0.5$. --- **Final answers:** - For $x^2 + 8x + 15 = 0$, $x = -3$ or $x = -5$. - For $2\omega + 2\omega^2$, simplified form is $2(\omega + \omega^2)$. - For $\cos 60^\circ$, the value is $\frac{1}{2}$ or $0.5$.