\( \frac{3x+1}{4}-\frac{7x-x^{2}}{10}=\frac{x^{2}-1}{} \)

To solve the equation \( \frac{3x+1}{4}-\frac{7x-x^{2}}{10}=\frac{x^{2}-1}{1} \), start by finding a common denominator for the fractions on the left side. The least common multiple of 4 and 10 is 20. Rewrite each term: \[ \frac{3x + 1}{4} = \frac{5(3x + 1)}{20} = \frac{15x + 5}{20} \] \[ \frac{7x - x^{2}}{10} = \frac{2(7x - x^{2})}{20} = \frac{14x - 2x^{2}}{20} \] Now substitute these into the equation: \[ \frac{15x + 5}{20} - \frac{14x - 2x^{2}}{20} = x^{2} - 1 \] Combine the left-side fractions: \[ \frac{15x + 5 - (14x - 2x^{2})}{20} = x^{2} - 1 \] This simplifies to: \[ \frac{15x + 5 - 14x + 2x^{2}}{20} = x^{2} - 1 \] \[ \frac{x + 5 + 2x^{2}}{20} = x^{2} - 1 \] Now, multiply both sides by 20 to eliminate the denominator: \[ x + 5 + 2x^{2} = 20(x^{2} - 1) \] Expanding the right side gives: \[ x + 5 + 2x^{2} = 20x^{2} - 20 \] Rearranging to bring all terms to one side yields: \[ 2x^{2} - 20x^{2} + x + 5 + 20 = 0 \] This simplifies to: \[ -18x^{2} + x + 25 = 0 \] Multiplying through by -1: \[ 18x^{2} - x - 25 = 0 \] Now use the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 18 \), \( b = -1 \), and \( c = -25 \): 1. Calculate the discriminant \( b^{2} - 4ac \): \[ (-1)^{2} - 4(18)(-25) = 1 + 1800 = 1801 \] 2. Now plug into the formula: \[ x = \frac{-(-1) \pm \sqrt{1801}}{2(18)} \] \[ x = \frac{1 \pm \sqrt{1801}}{36} \] Thus, the solutions are: \[ x = \frac{1 + \sqrt{1801}}{36} \quad \text{and} \quad x = \frac{1 - \sqrt{1801}}{36} \]