Solve the quadratic formula to solve \( 4 x^{2}-20 x+25=0 \). \[ x=\square \]

To solve the quadratic equation \( 4x^{2} - 20x + 25 = 0 \) using the quadratic formula, we start with \( ax^2 + bx + c = 0 \), where \( a = 4 \), \( b = -20 \), and \( c = 25 \). The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Calculating \( b^2 - 4ac \): \[ b^2 = (-20)^2 = 400 \] \[ 4ac = 4 \cdot 4 \cdot 25 = 400 \] \[ b^2 - 4ac = 400 - 400 = 0 \] Now we substitute these values into the quadratic formula: \[ x = \frac{20 \pm \sqrt{0}}{2 \cdot 4} \] \[ x = \frac{20 \pm 0}{8} = \frac{20}{8} = \frac{5}{2} \] Thus, the solution is: \[ x = \frac{5}{2} \] To add some more context, quadratic equations have been studied for centuries, evolving from ancient civilizations' need to solve problems related to land area and construction. They can describe a myriad of real-world phenomena such as projectile motion, economics, and even the shape of a bridge! Remember when using the quadratic formula, it's crucial to pay attention to the discriminant (\( b^2 - 4ac \)). If it’s positive, you'll find two distinct solutions, if it's zero, you'll get one repeated solution (as we did here), and if it's negative, you’ve got no real solutions—so don’t be surprised if a problem throws a curveball!