What are the solutions to the equation 0=128t-16t^2-192 ?

The solutions to the equation 0=128t-16t^2-192 are t=2,t=6

Find the zeros of 0=128t-16t^2-192

The zeros of 0=128t-16t^2-192 are t=2,t=6

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0 = 128 t - 16 t^{2} - 192

t = 2, t = 6

Solution steps

0 = 128 t - 16 t^{2} - 192

Switch sides

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

Write in the standard form

a x^{2} + b x + c = 0

- 16 t^{2} + 128 t - 192 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 16 ) ( - 192 )}{2 ( - 16 )}

12 8^{2} - 4 (- 16) (- 192) = 64

t_{1, 2} = \frac{- 128 \pm 64}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 128 + 64}{2 ( - 16 )}, t_{2} = \frac{- 128 - 64}{2 ( - 16 )}

t = \frac{- 128 + 64}{2 ( - 16 )} : 2

t = \frac{- 128 - 64}{2 ( - 16 )} : 6

The solutions to the quadratic equation are:

t = 2, t = 6

x (x - 12) + 35 = 0

4 e^{2} + e - 3 = 0

5 = x^{2} - x

x, y^{2} - 2 x^{2} = c

2 z^{2} - 4 z + 3 = 0

What are the solutions to the equation 0=128t-16t^2-192 ?
The solutions to the equation 0=128t-16t^2-192 are t=2,t=6
Find the zeros of 0=128t-16t^2-192
The zeros of 0=128t-16t^2-192 are t=2,t=6