What are the solutions to the equation -16t^2+16t+32=0 ?

The solutions to the equation -16t^2+16t+32=0 are t=-1,t=2

Find the zeros of -16t^2+16t+32=0

The zeros of -16t^2+16t+32=0 are t=-1,t=2

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- 16 t^{2} + 16 t + 32 = 0

t = - 1, t = 2

Solution steps

- 16 t^{2} + 16 t + 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) \cdot 32}{2 ( - 16 )}

1 6^{2} - 4 (- 16) \cdot 32 = 48

t_{1, 2} = \frac{- 16 \pm 48}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 16 + 48}{2 ( - 16 )}, t_{2} = \frac{- 16 - 48}{2 ( - 16 )}

t = \frac{- 16 + 48}{2 ( - 16 )} : - 1

t = \frac{- 16 - 48}{2 ( - 16 )} : 2

The solutions to the quadratic equation are:

t = - 1, t = 2

x^{2} + 4 x + 3 = 0

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

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

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

4 x^{2} + 48 = 0

What are the solutions to the equation -16t^2+16t+32=0 ?
The solutions to the equation -16t^2+16t+32=0 are t=-1,t=2
Find the zeros of -16t^2+16t+32=0
The zeros of -16t^2+16t+32=0 are t=-1,t=2