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Solve using the quadratic formula.\newline8m2−3m−9=08m^2 - 3m - 9 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinem=m = _____ or m=m = _____

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Q. Solve using the quadratic formula.\newline8m2−3m−9=08m^2 - 3m - 9 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinem=m = _____ or m=m = _____
  1. Quadratic Formula: The quadratic formula is given by m=−b±b2−4ac2am = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where aa, bb, and cc are the coefficients from the quadratic equation am2+bm+c=0am^2 + bm + c = 0. In this case, a=8a = 8, b=−3b = -3, and c=−9c = -9.
  2. Calculate Discriminant: First, calculate the discriminant, which is the part under the square root in the quadratic formula: b2−4acb^2 - 4ac. Discriminant = (−3)2−4(8)(−9)=9+288=297(-3)^2 - 4(8)(-9) = 9 + 288 = 297.
  3. Apply Quadratic Formula: Now, apply the quadratic formula using the values of aa, bb, and cc.m=−(−3)±2972×8m = \frac{-(-3) \pm \sqrt{297}}{2 \times 8}m=3±29716m = \frac{3 \pm \sqrt{297}}{16}
  4. Simplify 297\sqrt{297}: Since the discriminant is positive, there will be two real solutions. We need to simplify 297\sqrt{297}.\newline297\sqrt{297} is not a perfect square, so it cannot be simplified to an integer. We will leave it as 297\sqrt{297}.
  5. Calculate Solutions: Now, we will calculate the two solutions for mm. \newlinem1=3+29716m_1 = \frac{3 + \sqrt{297}}{16}\newlinem2=3−29716m_2 = \frac{3 - \sqrt{297}}{16}
  6. Approximate Solutions: These solutions can be approximated as decimals if necessary, but the question asks for integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinem1≈(3+17.23)/16≈20.23/16≈1.26m_1 \approx (3 + 17.23) / 16 \approx 20.23 / 16 \approx 1.26\newlinem2≈(3−17.23)/16≈−14.23/16≈−0.89m_2 \approx (3 - 17.23) / 16 \approx -14.23 / 16 \approx -0.89
  7. Round Decimal Solutions: Round the decimal solutions to the nearest hundredth. \newlinem1≈1.26m_1 \approx 1.26 (rounded to 1.261.26)\newlinem2≈−0.89m_2 \approx -0.89 (rounded to −0.89-0.89)

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