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

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

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

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

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

t = - 1, t = 2

Solution steps

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

Switch sides

- 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

[\frac{( x + 3 )}{( x + 8 )}] \geq 6

19 - 3 x \leq - (1 + 7 x)

- (a + 1) - 4 a \leq 2 a - 8

- \frac{6}{1}

2 a - 4

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