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Solve using the quadratic formula.\newline8z2+2z−9=08z^2 + 2z - 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.\newline8z2+2z−9=08z^2 + 2z - 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 of the quadratic equation.\newlineThe quadratic equation is in the form az2+bz+c=0az^2 + bz + c = 0. For the equation 8z2+2z−9=08z^2 + 2z - 9 = 0, the coefficients are:\newlinea = 88\newlineb = 22\newlinec = −9-9
  2. Substitute into Formula: Substitute the coefficients into the quadratic formula.\newlineThe quadratic formula is z=−b±b2−4ac2az = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Substituting the values of aa, bb, and cc, we get:\newlinez=−(2)±(2)2−4(8)(−9)2(8)z = \frac{-(2) \pm \sqrt{(2)^2 - 4(8)(-9)}}{2(8)}
  3. Simplify Square Root: Simplify under the square root.\newlineCalculate the discriminant (the expression under the square root):\newlineDiscriminant = (2)2−4(8)(−9)(2)^2 - 4(8)(-9)\newlineDiscriminant = 4+2884 + 288\newlineDiscriminant = 292292
  4. Continue with Formula: Continue with the quadratic formula.\newlineNow we have the discriminant, we can continue with the quadratic formula:\newlinez=−2±29216z = \frac{-2 \pm \sqrt{292}}{16}
  5. Simplify Solutions: Simplify the solutions.\newlineWe have two possible solutions for zz, corresponding to the '±\pm' in the formula:\newlinez=−2+29216z = \frac{-2 + \sqrt{292}}{16} or z=−2−29216z = \frac{-2 - \sqrt{292}}{16}
  6. Calculate Numerical Values: Calculate the numerical values and round to the nearest hundredth if necessary.\newlineFirst, calculate the square root of 292292:\newline292≈17.09\sqrt{292} \approx 17.09\newlineNow, substitute this value into the solutions:\newlinez=−2+17.0916z = \frac{-2 + 17.09}{16} or z=−2−17.0916z = \frac{-2 - 17.09}{16}\newlinez≈15.0916z \approx \frac{15.09}{16} or z≈−19.0916z \approx \frac{-19.09}{16}\newlinez≈0.94z \approx 0.94 or z≈−1.19z \approx -1.19

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