The equation \( 15 x^{2}-1 x-2=0 \) has solutions of the form \[ x=\frac{N \pm \sqrt{D}}{M} \] (A) Use the quadratic formula to solve this equation and find the appropriate integer values of \( N, M \), and \( D \). Do not worry about simplifying the \( \sqrt{D} \) yet in this part of the problem. \[ N=\square ; D=\square \] (B) Now simplify the radical and the resulting solutions. Enter your answers as a list of integers or reduced fractions, separated with commas. Example: \( -5 / 2,-3 / 4 \) \( x=\square \)

First, let's solve the equation \( 15x^2 - x - 2 = 0 \) using the quadratic formula: The quadratic formula states that the solutions for an equation of the form \( ax^2 + bx + c = 0 \) are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 15 \), \( b = -1 \), and \( c = -2 \). Now, let's calculate: 1. The discriminant \( D = b^2 - 4ac \): \[ D = (-1)^2 - 4(15)(-2) = 1 + 120 = 121 \] 2. Substitute \( a \), \( b \), and \( D \) into the formula: \[ x = \frac{-(-1) \pm \sqrt{121}}{2 \cdot 15} \] \[ x = \frac{1 \pm 11}{30} \] Thus, the values for \( N \), \( D \), and \( M \) are: \( N = 1, D = 121, M = 30 \) For part (B), let's simplify the radical and find the solutions: We have: \[ x = \frac{1 + 11}{30} \quad \text{and} \quad x = \frac{1 - 11}{30} \] Calculating these gives us: 1. \( x = \frac{12}{30} = \frac{2}{5} \) 2. \( x = \frac{-10}{30} = -\frac{1}{3} \) Therefore, the simplified solutions are: \( x = \frac{2}{5}, -\frac{1}{3} \)