What are the solutions to the equation 240=-16t^2+128t ?

The solutions to the equation 240=-16t^2+128t are t=3,t=5

Find the zeros of 240=-16t^2+128t

The zeros of 240=-16t^2+128t are t=3,t=5

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240 = - 16 t^{2} + 128 t

t = 3, t = 5

Solution steps

240 = - 16 t^{2} + 128 t

Switch sides

- 16 t^{2} + 128 t = 240

Move

240

to the left side

- 16 t^{2} + 128 t - 240 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 16 ) ( - 240 )}{2 ( - 16 )}

12 8^{2} - 4 (- 16) (- 240) = 32

t_{1, 2} = \frac{- 128 \pm 32}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 128 + 32}{2 ( - 16 )}, t_{2} = \frac{- 128 - 32}{2 ( - 16 )}

t = \frac{- 128 + 32}{2 ( - 16 )} : 3

t = \frac{- 128 - 32}{2 ( - 16 )} : 5

The solutions to the quadratic equation are:

t = 3, t = 5

(5 x - 2 y) (5 x + 2 y)

f (x) = 4 - x^{2}

1000 (\frac{1}{2})^{\frac{60}{5}}

\frac{110 - 100}{7}

6 + 2 x - 4 = x - 1

What are the solutions to the equation 240=-16t^2+128t ?
The solutions to the equation 240=-16t^2+128t are t=3,t=5
Find the zeros of 240=-16t^2+128t
The zeros of 240=-16t^2+128t are t=3,t=5