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

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

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

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

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0 = 128 t - 16 t^{2}

t = 0, t = 8

Solution steps

0 = 128 t - 16 t^{2}

Switch sides

128 t - 16 t^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 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

\frac{4 x + 7}{12} + x = \frac{3 x}{8}

\frac{8}{2 + 6 i}

6 4^{- \frac{4}{3}}

2 (x + 10)^{2} - 34 = - 26

7 r^{2} - 5 = 2 r + 9 r^{2}

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