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

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

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

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

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

t = 0, t = 8

Solution steps

- 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

p^{2} + 22 p - 3 = 0

\frac{3 a}{10} + \frac{3}{5} = \frac{a}{25} + \frac{1}{2}

\frac{9}{4} z + \frac{1}{2} = 2

0. 1^{2}

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