What are the solutions to the equation -16t^2+16t+480=0 ?

The solutions to the equation -16t^2+16t+480=0 are t=-5,t=6

Find the zeros of -16t^2+16t+480=0

The zeros of -16t^2+16t+480=0 are t=-5,t=6

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- 16 t^{2} + 16 t + 480 = 0

t = - 5, t = 6

Solution steps

- 16 t^{2} + 16 t + 480 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) \cdot 480}{2 ( - 16 )}

1 6^{2} - 4 (- 16) \cdot 480 = 176

t_{1, 2} = \frac{- 16 \pm 176}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 16 + 176}{2 ( - 16 )}, t_{2} = \frac{- 16 - 176}{2 ( - 16 )}

t = \frac{- 16 + 176}{2 ( - 16 )} : - 5

t = \frac{- 16 - 176}{2 ( - 16 )} : 6

The solutions to the quadratic equation are:

t = - 5, t = 6

x^{2} + 4 x - 2 = 0

(3 x^{2})^{- 1}

\frac{1}{3} + \frac{3}{4}

- 7 x > 35

- 6 x^{2} + 5 x + 6 = 0

What are the solutions to the equation -16t^2+16t+480=0 ?
The solutions to the equation -16t^2+16t+480=0 are t=-5,t=6
Find the zeros of -16t^2+16t+480=0
The zeros of -16t^2+16t+480=0 are t=-5,t=6