What are the solutions to the equation 0=-16t^2+16t+192 ?

The solutions to the equation 0=-16t^2+16t+192 are t=-3,t=4

Find the zeros of 0=-16t^2+16t+192

The zeros of 0=-16t^2+16t+192 are t=-3,t=4

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0 = - 16 t^{2} + 16 t + 192

t = - 3, t = 4

Solution steps

0 = - 16 t^{2} + 16 t + 192

Switch sides

- 16 t^{2} + 16 t + 192 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) \cdot 192}{2 ( - 16 )}

1 6^{2} - 4 (- 16) \cdot 192 = 112

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

Separate the solutions

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

t = \frac{- 16 + 112}{2 ( - 16 )} : - 3

t = \frac{- 16 - 112}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = - 3, t = 4

1 2^{2} + 1 2^{2} = c^{2}

1 < ∣ x - 2 ∣ < 4

e^{- \frac{1}{2} l n (x)}

3 x^{2} + 11 x + 8

(2 x - 1) (- 5 x + 10) \leq 0

What are the solutions to the equation 0=-16t^2+16t+192 ?
The solutions to the equation 0=-16t^2+16t+192 are t=-3,t=4
Find the zeros of 0=-16t^2+16t+192
The zeros of 0=-16t^2+16t+192 are t=-3,t=4