What are the solutions to the equation -16t^2+192t=0 ?

The solutions to the equation -16t^2+192t=0 are t=0,t=12

Find the zeros of -16t^2+192t=0

The zeros of -16t^2+192t=0 are t=0,t=12

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- 16 t^{2} + 192 t = 0

t = 0, t = 12

Solution steps

- 16 t^{2} + 192 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 192 \pm 19 2 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

19 2^{2} - 4 (- 16) \cdot 0 = 192

t_{1, 2} = \frac{- 192 \pm 192}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 192 + 192}{2 ( - 16 )}, t_{2} = \frac{- 192 - 192}{2 ( - 16 )}

t = \frac{- 192 + 192}{2 ( - 16 )} : 0

t = \frac{- 192 - 192}{2 ( - 16 )} : 12

The solutions to the quadratic equation are:

t = 0, t = 12

19 6^{\frac{1}{2}}

(1 + i) (1 + i)

12 x^{3} + 2 x^{2} - 30 x - 5

\frac{6 x}{x - 1} \leq \frac{x}{x + 3} + 6

\frac{x}{8} = 8

What are the solutions to the equation -16t^2+192t=0 ?
The solutions to the equation -16t^2+192t=0 are t=0,t=12
Find the zeros of -16t^2+192t=0
The zeros of -16t^2+192t=0 are t=0,t=12