What are the solutions to the equation 256=128t-16t^2 ?

The solutions to the equation 256=128t-16t^2 are t=4

Find the zeros of 256=128t-16t^2

The zeros of 256=128t-16t^2 are t=4

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256 = 128 t - 16 t^{2}

t = 4

Solution steps

256 = 128 t - 16 t^{2}

Switch sides

128 t - 16 t^{2} = 256

Move

256

to the left side

128 t - 16 t^{2} - 256 = 0

Write in the standard form

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

- 16 t^{2} + 128 t - 256 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 16 ) ( - 256 )}{2 ( - 16 )}

12 8^{2} - 4 (- 16) (- 256) = 0

t_{1, 2} = \frac{- 128 \pm 0}{2 ( - 16 )}

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

\frac{- 128}{2 ( - 16 )} = 4

t = 4

The solution to the quadratic equation is:

t = 4

0.2 (v - 5) = - 1

3 ∣ 5 x - 11 ∣ - 1 = 4

6 x^{2} - 2 x - 4

3 x y - 4 x^{2} + 6 y - 8 x

\frac{x}{2} - \frac{13}{6} \leq \frac{x}{3} + 1

What are the solutions to the equation 256=128t-16t^2 ?
The solutions to the equation 256=128t-16t^2 are t=4
Find the zeros of 256=128t-16t^2
The zeros of 256=128t-16t^2 are t=4