What are the solutions to the equation -4.9t^2+25t=0 ?

The solutions to the equation -4.9t^2+25t=0 are t=0,t= 250/49

Find the zeros of -4.9t^2+25t=0

The zeros of -4.9t^2+25t=0 are t=0,t= 250/49

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- 4.9 t^{2} + 25 t = 0

t = 0, t = \frac{250}{49}

Decimal Notation

t = 0, t = 5.10204 \dots

Solution steps

- 4.9 t^{2} + 25 t = 0

Multiply both sides by

10

- 49 t^{2} + 250 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 250 \pm 25 0 ^{2} - 4 ( - 49 ) \cdot 0}{2 ( - 49 )}

25 0^{2} - 4 (- 49) \cdot 0 = 250

t_{1, 2} = \frac{- 250 \pm 250}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 250 + 250}{2 ( - 49 )}, t_{2} = \frac{- 250 - 250}{2 ( - 49 )}

t = \frac{- 250 + 250}{2 ( - 49 )} : 0

t = \frac{- 250 - 250}{2 ( - 49 )} : \frac{250}{49}

The solutions to the quadratic equation are:

t = 0, t = \frac{250}{49}

- 4 x^{2} - 72 x - 68 = 0

16 (x^{2} - 2) + 13 = - 10

4 x^{2} - 9 x + 1 = 0

9 x^{2} - 9 x - 4 = 0

3 x^{2} + 12 x - 36 = 0

What are the solutions to the equation -4.9t^2+25t=0 ?
The solutions to the equation -4.9t^2+25t=0 are t=0,t= 250/49
Find the zeros of -4.9t^2+25t=0
The zeros of -4.9t^2+25t=0 are t=0,t= 250/49