What are the solutions to the equation 128=-16t^2+96t ?

The solutions to the equation 128=-16t^2+96t are t=2,t=4

Find the zeros of 128=-16t^2+96t

The zeros of 128=-16t^2+96t are t=2,t=4

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128 = - 16 t^{2} + 96 t

t = 2, t = 4

Solution steps

128 = - 16 t^{2} + 96 t

Switch sides

- 16 t^{2} + 96 t = 128

Move

128

to the left side

- 16 t^{2} + 96 t - 128 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 96 \pm 9 6 ^{2} - 4 ( - 16 ) ( - 128 )}{2 ( - 16 )}

9 6^{2} - 4 (- 16) (- 128) = 32

t_{1, 2} = \frac{- 96 \pm 32}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 96 + 32}{2 ( - 16 )}, t_{2} = \frac{- 96 - 32}{2 ( - 16 )}

t = \frac{- 96 + 32}{2 ( - 16 )} : 2

t = \frac{- 96 - 32}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = 2, t = 4

x^{2} + 48 x - 100 = 0

11 k^{2} = 20 + 3 k

(2 x - 5)^{2} = - 2 (6 x - 11)

- x^{2} - 10 x + (- 9) = 0

3 - x^{2} = x

What are the solutions to the equation 128=-16t^2+96t ?
The solutions to the equation 128=-16t^2+96t are t=2,t=4
Find the zeros of 128=-16t^2+96t
The zeros of 128=-16t^2+96t are t=2,t=4