What are the solutions to the equation 0=-4.9t^2+28t ?

The solutions to the equation 0=-4.9t^2+28t are t=0,t= 40/7

Find the zeros of 0=-4.9t^2+28t

The zeros of 0=-4.9t^2+28t are t=0,t= 40/7

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0 = - 4.9 t^{2} + 28 t

t = 0, t = \frac{40}{7}

Decimal Notation

t = 0, t = 5.71428 \dots

Solution steps

0 = - 4.9 t^{2} + 28 t

Multiply both sides by

10

0 = - 49 t^{2} + 280 t

Switch sides

- 49 t^{2} + 280 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 280 \pm 28 0 ^{2} - 4 ( - 49 ) \cdot 0}{2 ( - 49 )}

28 0^{2} - 4 (- 49) \cdot 0 = 280

t_{1, 2} = \frac{- 280 \pm 280}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 280 + 280}{2 ( - 49 )}, t_{2} = \frac{- 280 - 280}{2 ( - 49 )}

t = \frac{- 280 + 280}{2 ( - 49 )} : 0

t = \frac{- 280 - 280}{2 ( - 49 )} : \frac{40}{7}

The solutions to the quadratic equation are:

t = 0, t = \frac{40}{7}

12 x^{2} + 42 x - 294 = 0

x^{2} - x - 1 = 6 x - 1

x^{2} - \frac{1}{2} x = 0

11 x^{2} + 9 x = 0

u^{2} - u - 4 = 0

What are the solutions to the equation 0=-4.9t^2+28t ?
The solutions to the equation 0=-4.9t^2+28t are t=0,t= 40/7
Find the zeros of 0=-4.9t^2+28t
The zeros of 0=-4.9t^2+28t are t=0,t= 40/7