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Solve using the quadratic formula.\newline8v2−7v−8=08v^2 - 7v - 8 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinev=v = _____ or v=v = _____

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Q. Solve using the quadratic formula.\newline8v2−7v−8=08v^2 - 7v - 8 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinev=v = _____ or v=v = _____
  1. Identify values of aa, bb, cc: question_prompt: Solve the quadratic equation 8v2−7v−8=08v^2 − 7v − 8 = 0 using the quadratic formula.
  2. Plug values into formula: Identify values of aa, bb, and cc from the equation 8v2−7v−8=08v^2 − 7v − 8 = 0. Here, a=8a = 8, b=−7b = -7, and c=−8c = -8.
  3. Simplify square root: Plug aa, bb, and cc into the quadratic formula v=−b±b2−4ac2av = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. So, v=−(−7)±(−7)2−4⋅8⋅(−8)2⋅8v = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 8 \cdot (-8)}}{2 \cdot 8}.
  4. Calculate two solutions: Simplify inside the square root: 49+256\sqrt{49 + 256}. This is 305\sqrt{305}.
  5. First solution calculation: Now we have v=7±30516v = \frac{7 \pm \sqrt{305}}{16}. Calculate the two possible solutions for vv.
  6. Second solution calculation: First solution: v=7+30516v = \frac{7 + \sqrt{305}}{16}.
  7. Approximate square root: Second solution: v=7−30516v = \frac{7 - \sqrt{305}}{16}.
  8. Calculate approximate solutions: Approximate the square root of 305305 to the nearest hundredth: 305≈17.46\sqrt{305} \approx 17.46.
  9. First approximate solution: Calculate the approximate solutions: v≈(7+17.46)/16v \approx (7 + 17.46) / 16 and v≈(7−17.46)/16v \approx (7 - 17.46) / 16.
  10. Second approximate solution: First approximate solution: v≈24.4616v \approx \frac{24.46}{16}. This simplifies to v≈1.53v \approx 1.53.
  11. Second approximate solution: First approximate solution: v≈24.4616v \approx \frac{24.46}{16}. This simplifies to v≈1.53v \approx 1.53. Second approximate solution: v≈−10.4616v \approx \frac{-10.46}{16}. This simplifies to v≈−0.65v \approx -0.65.

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