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Solve using the quadratic formula.\newline5z2−7z−9=05z^2 - 7z - 9 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinez=z = _____ or z=z = _____

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Q. Solve using the quadratic formula.\newline5z2−7z−9=05z^2 - 7z - 9 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinez=z = _____ or z=z = _____
  1. Identify coefficients: Identify the coefficients aa, bb, and cc in the quadratic equation 5z2−7z−9=05z^2 − 7z − 9 = 0.a=5a = 5, b=−7b = -7, c=−9c = -9
  2. Write quadratic formula: Write down the quadratic formula: z=−b±b2−4ac2az = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
  3. Substitute values: Substitute the values of aa, bb, and cc into the quadratic formula.\newlinez=−(−7)±(−7)2−4⋅5⋅(−9)2⋅5z = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 5 \cdot (-9)}}{2 \cdot 5}
  4. Simplify expression: Simplify the expression inside the square root and the constants outside the square root.\newlinez=7±49+18010z = \frac{7 \pm \sqrt{49 + 180}}{10}\newlinez=7±22910z = \frac{7 \pm \sqrt{229}}{10}
  5. Calculate square root: Calculate the square root of 229229 to find the two possible solutions for zz.229 is approximately 15.13\sqrt{229} \text{ is approximately } 15.13 z=(7+15.13)10z = \frac{(7 + 15.13)}{10} or z=(7−15.13)10z = \frac{(7 - 15.13)}{10}
  6. Find z values: Simplify the expressions to find the values of zz.
    z=7+15.1310z = \frac{7 + 15.13}{10} or z=7−15.1310z = \frac{7 - 15.13}{10}
    z=22.1310z = \frac{22.13}{10} or z=−8.1310z = \frac{-8.13}{10}
    z≈2.21z \approx 2.21 or z≈−0.81z \approx -0.81

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