b) \( 2 x^{2}-8 x+2=0 \)

To solve the quadratic equation \(2x^{2} - 8x + 2 = 0\), you can simplify it by dividing every term by 2, resulting in \(x^{2} - 4x + 1 = 0\). Now, use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = 1\). Plugging in these values gives you \(x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}\). Therefore, the roots of the equation are \(x = 2 + \sqrt{3}\) and \(x = 2 - \sqrt{3}\). The quadratic formula can be like a magical spell for solving any quadratic equation! Just remember that \(b^2 - 4ac\) (the discriminant) tells you about the nature of the roots: if it’s positive, two distinct real solutions; if it’s zero, one double root; and if negative, no real roots, just complex ones. So, mastering this formula can make you a wizard in the realm of polynomials! You can further explore quadratics through real-world applications like projectile motion or finance. For instance, the path of a thrown ball can be modeled by a quadratic equation. Excitingly, understanding the vertex and roots helps predict the highest point of the ball's trajectory and when it will hit the ground. Dive into any physics or finance book that discusses parabolas to see how these equations play a crucial role in the real world!