What are the solutions to the equation -4.9t^2+13t+18=0 ?

The solutions to the equation -4.9t^2+13t+18=0 are t=-(-65+sqrt(13045))/(49),t=(65+sqrt(13045))/(49)

Find the zeros of -4.9t^2+13t+18=0

The zeros of -4.9t^2+13t+18=0 are t=-(-65+sqrt(13045))/(49),t=(65+sqrt(13045))/(49)

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- 4.9 t^{2} + 13 t + 18 = 0

t = - \frac{- 65 + 13045}{49}, t = \frac{65 + 13045}{49}

Decimal Notation

t = - 1.00438 \dots, t = 3.65744 \dots

Solution steps

- 4.9 t^{2} + 13 t + 18 = 0

Multiply both sides by

10

- 49 t^{2} + 130 t + 180 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 130 \pm 13 0 ^{2} - 4 ( - 49 ) \cdot 180}{2 ( - 49 )}

13 0^{2} - 4 (- 49) \cdot 180 = 213045

t_{1, 2} = \frac{- 130 \pm 2 13045}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 130 + 2 13045}{2 ( - 49 )}, t_{2} = \frac{- 130 - 2 13045}{2 ( - 49 )}

t = \frac{- 130 + 2 13045}{2 ( - 49 )} : - \frac{- 65 + 13045}{49}

t = \frac{- 130 - 2 13045}{2 ( - 49 )} : \frac{65 + 13045}{49}

The solutions to the quadratic equation are:

t = - \frac{- 65 + 13045}{49}, t = \frac{65 + 13045}{49}

5 x^{2} - 2 = - 88

25 x^{2} + 1 = 20 x

\frac{1}{2} c^{2} + 3 c = 1

t^{2} - t + 6 = 0

x^{2} + 2 x - 9 = 0

What are the solutions to the equation -4.9t^2+13t+18=0 ?
The solutions to the equation -4.9t^2+13t+18=0 are t=-(-65+sqrt(13045))/(49),t=(65+sqrt(13045))/(49)
Find the zeros of -4.9t^2+13t+18=0
The zeros of -4.9t^2+13t+18=0 are t=-(-65+sqrt(13045))/(49),t=(65+sqrt(13045))/(49)