What are the solutions to the equation -2f^2+32f-22=0 ?

The solutions to the equation -2f^2+32f-22=0 are f=8-sqrt(53),f=8+sqrt(53)

Find the zeros of -2f^2+32f-22=0

The zeros of -2f^2+32f-22=0 are f=8-sqrt(53),f=8+sqrt(53)

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- 2 f^{2} + 32 f - 22 = 0

f = 8 - 53, f = 8 + 53

Decimal Notation

f = 0.71989 \dots, f = 15.28010 \dots

Solution steps

- 2 f^{2} + 32 f - 22 = 0

Solve with the quadratic formula

f_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 2 ) ( - 22 )}{2 ( - 2 )}

3 2^{2} - 4 (- 2) (- 22) = 453

f_{1, 2} = \frac{- 32 \pm 4 53}{2 ( - 2 )}

Separate the solutions

f_{1} = \frac{- 32 + 4 53}{2 ( - 2 )}, f_{2} = \frac{- 32 - 4 53}{2 ( - 2 )}

f = \frac{- 32 + 4 53}{2 ( - 2 )} : 8 - 53

f = \frac{- 32 - 4 53}{2 ( - 2 )} : 8 + 53

The solutions to the quadratic equation are:

f = 8 - 53, f = 8 + 53

x^{2} + 7 x - 5 = 0

\frac{11 k - 15}{3} > \frac{5 - 2 k}{4}

∣ x ∣ - 2 = 3

(9 x^{8})^{\frac{1}{2}}

- 5 x^{2} = - 25

What are the solutions to the equation -2f^2+32f-22=0 ?
The solutions to the equation -2f^2+32f-22=0 are f=8-sqrt(53),f=8+sqrt(53)
Find the zeros of -2f^2+32f-22=0
The zeros of -2f^2+32f-22=0 are f=8-sqrt(53),f=8+sqrt(53)