To solve the quadratic equation \(21f^2 - 28f - 105 = 0\), we can use the quadratic formula:
\[ f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 21\), \(b = -28\), and \(c = -105\).
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 = (-28)^2 = 784 \] \[ 4ac = 4 \cdot 21 \cdot (-105) = -8820 \] \[ b^2 - 4ac = 784 - (-8820) = 784 + 8820 = 9604 \]
Next, we find the square root of the discriminant:
\[ \sqrt{9604} = 98 \]
Now, we use the quadratic formula:
\[ f = \frac{-(-28) \pm 98}{2 \cdot 21} \] \[ f = \frac{28 \pm 98}{42} \]
This gives us two cases:
- First Case: \(f = \frac{28 + 98}{42} = \frac{126}{42} = 3\)
- Second Case: \(f = \frac{28 - 98}{42} = \frac{-70}{42} = \frac{-5}{3}\)
Thus, the solutions for \(f\) are:
\[ \boxed{3, -\frac{5}{3}} \]
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