What are the solutions to the equation f^2+28f-47=0 ?

The solutions to the equation f^2+28f-47=0 are f=-14+9sqrt(3),f=-14-9sqrt(3)

Find the zeros of f^2+28f-47=0

The zeros of f^2+28f-47=0 are f=-14+9sqrt(3),f=-14-9sqrt(3)

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f^{2} + 28 f - 47 = 0

f = - 14 + 93, f = - 14 - 93

Solution steps

f^{2} + 28 f - 47 = 0

Solve with the quadratic formula

f_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 1 \cdot ( - 47 )}{2 \cdot 1}

2 8^{2} - 4 \cdot 1 \cdot (- 47) = 183

f_{1, 2} = \frac{- 28 \pm 18 3}{2 \cdot 1}

Separate the solutions

f_{1} = \frac{- 28 + 18 3}{2 \cdot 1}, f_{2} = \frac{- 28 - 18 3}{2 \cdot 1}

f = \frac{- 28 + 18 3}{2 \cdot 1} : - 14 + 93

f = \frac{- 28 - 18 3}{2 \cdot 1} : - 14 - 93

The solutions to the quadratic equation are:

f = - 14 + 93, f = - 14 - 93

0 = t^{2} - 190 t - 6000

2 ∣ 6 - 2 x ∣ + 6 = 10

x^{\frac{1}{2}} \cdot x^{3}

20 t + \frac{1}{2} (9.8) t^{2} = 12 t + \frac{1}{2} (- 9.8) (t - 3)^{2}

40 = t^{2} + 3 t

What are the solutions to the equation f^2+28f-47=0 ?
The solutions to the equation f^2+28f-47=0 are f=-14+9sqrt(3),f=-14-9sqrt(3)
Find the zeros of f^2+28f-47=0
The zeros of f^2+28f-47=0 are f=-14+9sqrt(3),f=-14-9sqrt(3)