What are the solutions to the equation -16t^2+96t+112=0 ?

The solutions to the equation -16t^2+96t+112=0 are t=-1,t=7

Find the zeros of -16t^2+96t+112=0

The zeros of -16t^2+96t+112=0 are t=-1,t=7

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- 16 t^{2} + 96 t + 112 = 0

t = - 1, t = 7

Solution steps

- 16 t^{2} + 96 t + 112 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 96 \pm 9 6 ^{2} - 4 ( - 16 ) \cdot 112}{2 ( - 16 )}

9 6^{2} - 4 (- 16) \cdot 112 = 128

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

Separate the solutions

t_{1} = \frac{- 96 + 128}{2 ( - 16 )}, t_{2} = \frac{- 96 - 128}{2 ( - 16 )}

t = \frac{- 96 + 128}{2 ( - 16 )} : - 1

t = \frac{- 96 - 128}{2 ( - 16 )} : 7

The solutions to the quadratic equation are:

t = - 1, t = 7

10.4 = 19 t - 4.9 t^{2}

2 (2 x - 1) + 2 x = 6 (x - 1)

1 - \frac{2}{3} + \frac{1}{5}

81 x^{2} + 144 x y + 64 y^{2}

pq - 3 r + p r - 3 q

What are the solutions to the equation -16t^2+96t+112=0 ?
The solutions to the equation -16t^2+96t+112=0 are t=-1,t=7
Find the zeros of -16t^2+96t+112=0
The zeros of -16t^2+96t+112=0 are t=-1,t=7