What are the solutions to the equation -14s^2+440s-3300=0 ?

The solutions to the equation -14s^2+440s-3300=0 are s=(5(22-sqrt(22)))/7 ,s=(5(22+sqrt(22)))/7

Find the zeros of -14s^2+440s-3300=0

The zeros of -14s^2+440s-3300=0 are s=(5(22-sqrt(22)))/7 ,s=(5(22+sqrt(22)))/7

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- 14 s^{2} + 440 s - 3300 = 0

s = \frac{5 ( 22 - 22 )}{7}, s = \frac{5 ( 22 + 22 )}{7}

Decimal Notation

s = 12.36398 \dots, s = 19.06458 \dots

Solution steps

- 14 s^{2} + 440 s - 3300 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 440 \pm 44 0 ^{2} - 4 ( - 14 ) ( - 3300 )}{2 ( - 14 )}

44 0^{2} - 4 (- 14) (- 3300) = 2022

s_{1, 2} = \frac{- 440 \pm 20 22}{2 ( - 14 )}

Separate the solutions

s_{1} = \frac{- 440 + 20 22}{2 ( - 14 )}, s_{2} = \frac{- 440 - 20 22}{2 ( - 14 )}

s = \frac{- 440 + 20 22}{2 ( - 14 )} : \frac{5 ( 22 - 22 )}{7}

s = \frac{- 440 - 20 22}{2 ( - 14 )} : \frac{5 ( 22 + 22 )}{7}

The solutions to the quadratic equation are:

s = \frac{5 ( 22 - 22 )}{7}, s = \frac{5 ( 22 + 22 )}{7}

8 x^{2} - 8 x + 21 = 0

3 x^{2} - 2 x - 37 = 0

4 p^{2} - 7 p + 6 = 5

0 = 1 - x - x^{2}

y, 3 x^{2} y + 48 x = 0

What are the solutions to the equation -14s^2+440s-3300=0 ?
The solutions to the equation -14s^2+440s-3300=0 are s=(5(22-sqrt(22)))/7 ,s=(5(22+sqrt(22)))/7
Find the zeros of -14s^2+440s-3300=0
The zeros of -14s^2+440s-3300=0 are s=(5(22-sqrt(22)))/7 ,s=(5(22+sqrt(22)))/7