What are the solutions to the equation 0=-16t^2+96t+128 ?

The solutions to the equation 0=-16t^2+96t+128 are t=3-sqrt(17),t=3+sqrt(17)

Find the zeros of 0=-16t^2+96t+128

The zeros of 0=-16t^2+96t+128 are t=3-sqrt(17),t=3+sqrt(17)

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

t = 3 - 17, t = 3 + 17

Decimal Notation

t = - 1.12310 \dots, t = 7.12310 \dots

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

9 6^{2} - 4 (- 16) \cdot 128 = 3217

t_{1, 2} = \frac{- 96 \pm 32 17}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 96 + 32 17}{2 ( - 16 )}, t_{2} = \frac{- 96 - 32 17}{2 ( - 16 )}

t = \frac{- 96 + 32 17}{2 ( - 16 )} : 3 - 17

t = \frac{- 96 - 32 17}{2 ( - 16 )} : 3 + 17

The solutions to the quadratic equation are:

t = 3 - 17, t = 3 + 17

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What are the solutions to the equation 0=-16t^2+96t+128 ?
The solutions to the equation 0=-16t^2+96t+128 are t=3-sqrt(17),t=3+sqrt(17)
Find the zeros of 0=-16t^2+96t+128
The zeros of 0=-16t^2+96t+128 are t=3-sqrt(17),t=3+sqrt(17)