What are the solutions to the equation 192=-16t^2+128t ?

The solutions to the equation 192=-16t^2+128t are t=2,t=6

Find the zeros of 192=-16t^2+128t

The zeros of 192=-16t^2+128t are t=2,t=6

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192 = - 16 t^{2} + 128 t

t = 2, t = 6

Solution steps

192 = - 16 t^{2} + 128 t

Switch sides

- 16 t^{2} + 128 t = 192

Move

192

to the left side

- 16 t^{2} + 128 t - 192 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 16 ) ( - 192 )}{2 ( - 16 )}

12 8^{2} - 4 (- 16) (- 192) = 64

t_{1, 2} = \frac{- 128 \pm 64}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 128 + 64}{2 ( - 16 )}, t_{2} = \frac{- 128 - 64}{2 ( - 16 )}

t = \frac{- 128 + 64}{2 ( - 16 )} : 2

t = \frac{- 128 - 64}{2 ( - 16 )} : 6

The solutions to the quadratic equation are:

t = 2, t = 6

- 1 + x < - 10

- 0.9 + 0.6 t \leq 0.1 t + 0.5

\frac{4 ( x + 6 )}{8} = 3 x - 2

\frac{π}{3} + \frac{π}{6}

- 2 (4 x y^{5})^{2}

What are the solutions to the equation 192=-16t^2+128t ?
The solutions to the equation 192=-16t^2+128t are t=2,t=6
Find the zeros of 192=-16t^2+128t
The zeros of 192=-16t^2+128t are t=2,t=6