Solve for \( v \). \( \frac{2}{v^{2}-25}-\frac{3}{v-5}=\frac{1}{v+5} \) If there is more than one solution, separate them with com If there is no solution, click on "No solution".

To solve the equation \( \frac{2}{v^{2}-25}-\frac{3}{v-5}=\frac{1}{v+5} \), we start by factoring the denominator of the first term: \[ v^{2} - 25 = (v - 5)(v + 5) \] The equation then becomes: \[ \frac{2}{(v-5)(v+5)} - \frac{3}{v-5} = \frac{1}{v+5} \] Next, we find a common denominator for the left-hand side, which is \((v-5)(v+5)\): \[ \frac{2 - 3(v+5)}{(v-5)(v+5)} = \frac{1}{v+5} \] This simplifies to: \[ \frac{2 - 3v - 15}{(v-5)(v+5)} = \frac{1}{v+5} \] Combine like terms in the numerator: \[ \frac{-3v - 13}{(v-5)(v+5)} = \frac{1}{v+5} \] Cross-multiply to eliminate the fractions: \[ (-3v - 13)(v+5) = (v-5) \] Expanding both sides results in: \[ -3v^2 - 15v - 13v - 65 = v - 5 \] Combining like terms gives: \[ -3v^2 - 28v - 65 = v - 5 \] Rearranging leads to: \[ -3v^2 - 29v - 60 = 0 \] To make calculations simpler, we can multiply the entire equation by -1: \[ 3v^2 + 29v + 60 = 0 \] Now we can use the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = 29, c = 60 \): Calculating the discriminant: \[ b^2 - 4ac = 29^2 - 4 \cdot 3 \cdot 60 = 841 - 720 = 121 \] Now, substituting back into the quadratic formula: \[ v = \frac{-29 \pm \sqrt{121}}{2 \cdot 3} = \frac{-29 \pm 11}{6} \] This gives us two potential solutions: 1. \( v = \frac{-29 + 11}{6} = \frac{-18}{6} = -3 \) 2. \( v = \frac{-29 - 11}{6} = \frac{-40}{6} = -\frac{20}{3} \) Thus, the solutions for \( v \) are: \[ v = -3, -\frac{20}{3} \]