1. We are asked to solve the quadratic equation $4x^2 + ? = 0$ (assuming the general form $ax^2 + bx + c = 0$). 2. To proceed, I'll demonstrate solving a quadratic equation by two methods: completing the square and using the quadratic formula. ### Method 1: Completing the Square 3. Consider a quadratic equation in standard form: $$ax^2 + bx + c = 0$$. For illustration, let's solve $$x^2 + 6x + 5 = 0$$. 4. Divide both sides by $a=1$ (here $a=1$, so no change). 5. Move the constant term to the right side: $$x^2 + 6x = -5$$. 6. To complete the square, add $$\left(\frac{b}{2a}\right)^2 = \left(\frac{6}{2}\right)^2 = 3^2 = 9$$ to both sides: $$x^2 + 6x + 9 = -5 + 9$$. 7. Left side becomes a perfect square: $$\left(x + 3\right)^2 = 4$$. 8. Take square roots: $$x + 3 = \pm 2$$. 9. Solve for $x$: - $x = -3 + 2 = -1$ - $x = -3 - 2 = -5$ ### Method 2: Quadratic Formula 10. The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 11. Using the same equation $x^2 + 6x + 5 = 0$, identify $a=1$, $b=6$, $c=5$. 12. Compute the discriminant: $$b^2 - 4ac = 6^2 - 4(1)(5) = 36 - 20 = 16$$. 13. Calculate roots: $$x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2}$$. 14. Thus, the solutions are: - $x = \frac{-6 + 4}{2} = \frac{-2}{2} = -1$ - $x = \frac{-6 - 4}{2} = \frac{-10}{2} = -5$ Final answer: The quadratic equation $x^2 + 6x + 5 = 0$ has roots $x = -1$ and $x = -5$ by both methods.