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Solve for ff.\newlinef2+12f−13=0f^2 + 12f - 13 = 0\newlineWrite each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.\newlinef=f = ____

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Q. Solve for ff.\newlinef2+12f−13=0f^2 + 12f - 13 = 0\newlineWrite each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.\newlinef=f = ____
  1. Identify the quadratic equation: Identify the quadratic equation.\newlineThe given equation is f2+12f−13=0f^2 + 12f - 13 = 0, which is a quadratic equation in the standard form af2+bf+c=0af^2 + bf + c = 0, where a=1a = 1, b=12b = 12, and c=−13c = -13.
  2. Use the quadratic formula: Use the quadratic formula to solve for ff. The quadratic formula is f=−b±b2−4ac2af = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. For our equation, a=1a = 1, b=12b = 12, and c=−13c = -13.
  3. Calculate the discriminant: Calculate the discriminant.\newlineThe discriminant is the part of the quadratic formula under the square root: b2−4acb^2 - 4ac. Let's calculate it:\newlineDiscriminant = (12)2−4(1)(−13)=144+52=196(12)^2 - 4(1)(-13) = 144 + 52 = 196.
  4. Two real solutions: Since the discriminant is positive, there are two real solutions.\newlineNow we can find the two solutions using the quadratic formula:\newlinef=−12±1962×1f = \frac{-12 \pm \sqrt{196}}{2 \times 1}.
  5. Calculate the two solutions: Calculate the two solutions.\newlineFirst solution:\newlinef=−12+1962=−12+142=22=1f = \frac{-12 + \sqrt{196}}{2} = \frac{-12 + 14}{2} = \frac{2}{2} = 1.\newlineSecond solution:\newlinef=−12−1962=−12−142=−262=−13f = \frac{-12 - \sqrt{196}}{2} = \frac{-12 - 14}{2} = \frac{-26}{2} = -13.

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