What are the solutions to the equation -16t^2+80t+96=0 ?

The solutions to the equation -16t^2+80t+96=0 are t=-1,t=6

Find the zeros of -16t^2+80t+96=0

The zeros of -16t^2+80t+96=0 are t=-1,t=6

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- 16 t^{2} + 80 t + 96 = 0

t = - 1, t = 6

Solution steps

- 16 t^{2} + 80 t + 96 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 16 ) \cdot 96}{2 ( - 16 )}

8 0^{2} - 4 (- 16) \cdot 96 = 112

t_{1, 2} = \frac{- 80 \pm 112}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 80 + 112}{2 ( - 16 )}, t_{2} = \frac{- 80 - 112}{2 ( - 16 )}

t = \frac{- 80 + 112}{2 ( - 16 )} : - 1

t = \frac{- 80 - 112}{2 ( - 16 )} : 6

The solutions to the quadratic equation are:

t = - 1, t = 6

x^{2} + 10 = - 11 x

x^{2} = 7 x + 18

6 x^{2} - 6 = 0

\frac{t}{5} - 13 = - 3

x^{2} - 2 x + 3 x - 6

What are the solutions to the equation -16t^2+80t+96=0 ?
The solutions to the equation -16t^2+80t+96=0 are t=-1,t=6
Find the zeros of -16t^2+80t+96=0
The zeros of -16t^2+80t+96=0 are t=-1,t=6