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

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

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

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

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192 = 128 t - 16 t^{2}

t = 2, t = 6

Solution steps

192 = 128 t - 16 t^{2}

Switch sides

128 t - 16 t^{2} = 192

Move

192

to the left side

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^{2} - 16 x - 65

4. 5^{2} + 3. 5^{2} = x^{2}

5 x^{2} - 15 x + 50 = 0

2 x^{2} + 7 x - 4 = 0

6 - 5 a^{2} = - 6 a + 4

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