What are the solutions to the equation 0=-16t^2+128t ?

The solutions to the equation 0=-16t^2+128t are t=0,t=8

Find the zeros of 0=-16t^2+128t

The zeros of 0=-16t^2+128t are t=0,t=8

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0 = - 16 t^{2} + 128 t

t = 0, t = 8

Solution steps

0 = - 16 t^{2} + 128 t

Switch sides

- 16 t^{2} + 128 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

12 8^{2} - 4 (- 16) \cdot 0 = 128

t_{1, 2} = \frac{- 128 \pm 128}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 128 + 128}{2 ( - 16 )}, t_{2} = \frac{- 128 - 128}{2 ( - 16 )}

t = \frac{- 128 + 128}{2 ( - 16 )} : 0

t = \frac{- 128 - 128}{2 ( - 16 )} : 8

The solutions to the quadratic equation are:

t = 0, t = 8

x - 6 - 10 - x \geq 1

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

8 - 4 (2 - x) \leq - 2 x

1 + \frac{2}{x + 1} - \frac{2}{x} \leq 0

1 - \frac{x - 3}{2} \geq \frac{x}{2} + 5

What are the solutions to the equation 0=-16t^2+128t ?
The solutions to the equation 0=-16t^2+128t are t=0,t=8
Find the zeros of 0=-16t^2+128t
The zeros of 0=-16t^2+128t are t=0,t=8