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Solve using the quadratic formula.\newline5z2z7=05z^2 - z - 7 = 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.\newline5z2z7=05z^2 - z - 7 = 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: To solve the quadratic equation 5z2z7=05z^2 - z - 7 = 0 using the quadratic formula, we first identify the coefficients aa, bb, and cc from the standard form of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. Here, a=5a = 5, b=1b = -1, and c=7c = -7.
  2. Apply quadratic formula: The quadratic formula is given by z=b±b24ac2az = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. We will substitute the values of aa, bb, and cc into this formula to find the solutions for zz.
  3. Calculate discriminant: First, calculate the discriminant, which is the part under the square root in the quadratic formula: b24acb^2 - 4ac. For our equation, the discriminant is (1)24(5)(7)=1+140=141(-1)^2 - 4(5)(-7) = 1 + 140 = 141.
  4. Plug values into formula: Now, we can plug the values into the quadratic formula: z=(1)±1412×5z = \frac{-(-1) \pm \sqrt{141}}{2 \times 5}. This simplifies to z=1±14110z = \frac{1 \pm \sqrt{141}}{10}.
  5. Calculate first solution: Since the discriminant is positive, we have two real and distinct solutions. We will calculate each solution separately, starting with the addition of the square root.\newlinez=1+141101+11.8741012.874101.29z = \frac{1 + \sqrt{141}}{10} \approx \frac{1 + 11.874}{10} \approx \frac{12.874}{10} \approx 1.29 when rounded to the nearest hundredth.
  6. Calculate second solution: Next, we calculate the solution with the subtraction of the square root. z=114110111.8741010.874101.09z = \frac{1 - \sqrt{141}}{10} \approx \frac{1 - 11.874}{10} \approx \frac{-10.874}{10} \approx -1.09 when rounded to the nearest hundredth.

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