What are the solutions to the equation 0=2+24.5t-4.9t^2 ?

The solutions to the equation 0=2+24.5t-4.9t^2 are t=-(-35+3sqrt(145))/(14),t=(35+3sqrt(145))/(14)

Find the zeros of 0=2+24.5t-4.9t^2

The zeros of 0=2+24.5t-4.9t^2 are t=-(-35+3sqrt(145))/(14),t=(35+3sqrt(145))/(14)

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0 = 2 + 24.5 t - 4.9 t^{2}

t = - \frac{- 35 + 3 145}{14}, t = \frac{35 + 3 145}{14}

Decimal Notation

t = - 0.08034 \dots, t = 5.08034 \dots

Solution steps

0 = 2 + 24.5 t - 4.9 t^{2}

Multiply both sides by

10

0 = 20 + 245 t - 49 t^{2}

Switch sides

20 + 245 t - 49 t^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 49 t^{2} + 245 t + 20 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 245 \pm 24 5 ^{2} - 4 ( - 49 ) \cdot 20}{2 ( - 49 )}

24 5^{2} - 4 (- 49) \cdot 20 = 21145

t_{1, 2} = \frac{- 245 \pm 21 145}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 245 + 21 145}{2 ( - 49 )}, t_{2} = \frac{- 245 - 21 145}{2 ( - 49 )}

t = \frac{- 245 + 21 145}{2 ( - 49 )} : - \frac{- 35 + 3 145}{14}

t = \frac{- 245 - 21 145}{2 ( - 49 )} : \frac{35 + 3 145}{14}

The solutions to the quadratic equation are:

t = - \frac{- 35 + 3 145}{14}, t = \frac{35 + 3 145}{14}

x^{2} + 8 x = 65

5 x^{2} - 18 x + 9 \leq 0

p^{2} + 3 p - 54

(n + 1) (n + 1)!

9 u^{2} - 80 u - 9

What are the solutions to the equation 0=2+24.5t-4.9t^2 ?
The solutions to the equation 0=2+24.5t-4.9t^2 are t=-(-35+3sqrt(145))/(14),t=(35+3sqrt(145))/(14)
Find the zeros of 0=2+24.5t-4.9t^2
The zeros of 0=2+24.5t-4.9t^2 are t=-(-35+3sqrt(145))/(14),t=(35+3sqrt(145))/(14)