What are the solutions to the equation 0=-4.9t^2+32t ?

The solutions to the equation 0=-4.9t^2+32t are t=0,t= 320/49

Find the zeros of 0=-4.9t^2+32t

The zeros of 0=-4.9t^2+32t are t=0,t= 320/49

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0 = - 4.9 t^{2} + 32 t

t = 0, t = \frac{320}{49}

Decimal Notation

t = 0, t = 6.53061 \dots

Solution steps

0 = - 4.9 t^{2} + 32 t

Multiply both sides by

10

0 = - 49 t^{2} + 320 t

Switch sides

- 49 t^{2} + 320 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 320 \pm 32 0 ^{2} - 4 ( - 49 ) \cdot 0}{2 ( - 49 )}

32 0^{2} - 4 (- 49) \cdot 0 = 320

t_{1, 2} = \frac{- 320 \pm 320}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 320 + 320}{2 ( - 49 )}, t_{2} = \frac{- 320 - 320}{2 ( - 49 )}

t = \frac{- 320 + 320}{2 ( - 49 )} : 0

t = \frac{- 320 - 320}{2 ( - 49 )} : \frac{320}{49}

The solutions to the quadratic equation are:

t = 0, t = \frac{320}{49}

- x^{2} + 13 x - 40 = 0

x = 2 x^{2} + 3 x - 1

3 x^{2} + 15 = 18 x

2 (x + 1)^{2} - 3 = 0

2 x^{2} - 3 x - 15 = 0

What are the solutions to the equation 0=-4.9t^2+32t ?
The solutions to the equation 0=-4.9t^2+32t are t=0,t= 320/49
Find the zeros of 0=-4.9t^2+32t
The zeros of 0=-4.9t^2+32t are t=0,t= 320/49