What are the solutions to the equation 0=-16t^2+112t+128 ?

The solutions to the equation 0=-16t^2+112t+128 are t=-1,t=8

Find the zeros of 0=-16t^2+112t+128

The zeros of 0=-16t^2+112t+128 are t=-1,t=8

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0 = - 16 t^{2} + 112 t + 128

t = - 1, t = 8

Solution steps

0 = - 16 t^{2} + 112 t + 128

Switch sides

- 16 t^{2} + 112 t + 128 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 112 \pm 11 2 ^{2} - 4 ( - 16 ) \cdot 128}{2 ( - 16 )}

11 2^{2} - 4 (- 16) \cdot 128 = 144

t_{1, 2} = \frac{- 112 \pm 144}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 112 + 144}{2 ( - 16 )}, t_{2} = \frac{- 112 - 144}{2 ( - 16 )}

t = \frac{- 112 + 144}{2 ( - 16 )} : - 1

t = \frac{- 112 - 144}{2 ( - 16 )} : 8

The solutions to the quadratic equation are:

t = - 1, t = 8

y^{2} + 5 y - 24 = 0

2 (25 - 3 x^{2}) = 0

x^{2} = 56

x^{2} = 53

0 = - x^{2} + 6 x - 8

What are the solutions to the equation 0=-16t^2+112t+128 ?
The solutions to the equation 0=-16t^2+112t+128 are t=-1,t=8
Find the zeros of 0=-16t^2+112t+128
The zeros of 0=-16t^2+112t+128 are t=-1,t=8