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

The solutions to the equation -16t^2+16t+96=0 are t=-2,t=3

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

The zeros of -16t^2+16t+96=0 are t=-2,t=3

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

t = - 2, t = 3

Solution steps

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

Solve with the quadratic formula

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

1 6^{2} - 4 (- 16) \cdot 96 = 80

t_{1, 2} = \frac{- 16 \pm 80}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 16 + 80}{2 ( - 16 )}, t_{2} = \frac{- 16 - 80}{2 ( - 16 )}

t = \frac{- 16 + 80}{2 ( - 16 )} : - 2

t = \frac{- 16 - 80}{2 ( - 16 )} : 3

The solutions to the quadratic equation are:

t = - 2, t = 3

g (n) = - 3 n^{2} + 7 n + 1

x, f = (y - 3 x)^{2}

t^{2} - 50 = 0

x, 2 x^{2} - x - 1 = 0

- x^{2} - 5 x + 6 = 2 x^{2} + 4 x - 11

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