Solve the equation for all real solutions in simplest form. \[ 4 d^{2}-19 d+20=0 \]

To solve the equation \( 4 d^{2}-19 d+20=0 \), we can either factor or use the quadratic formula. Here, the quadratic formula is the most straightforward method to ensure we find all solutions. The quadratic formula is given by: \[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our equation, \( a = 4 \), \( b = -19 \), and \( c = 20 \). Plugging these values into the formula: 1. Calculate the discriminant: \[ b^2 - 4ac = (-19)^2 - 4 \cdot 4 \cdot 20 = 361 - 320 = 41 \] 2. Substitute into the quadratic formula: \[ d = \frac{-(-19) \pm \sqrt{41}}{2 \cdot 4} = \frac{19 \pm \sqrt{41}}{8} \] Thus, there are two real solutions: \[ d = \frac{19 + \sqrt{41}}{8} \quad \text{and} \quad d = \frac{19 - \sqrt{41}}{8} \] These values represent the simplest form of the solutions for the given quadratic equation.