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

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

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

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

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240 = 128 t - 16 t^{2}

t = 3, t = 5

Solution steps

240 = 128 t - 16 t^{2}

Switch sides

128 t - 16 t^{2} = 240

Move

240

to the left side

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

Write in the standard form

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

- 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

y, y^{2} = 4 x^{2} + x^{3} - 36 - 9 x

x, 81 x^{2} + 72 x + 28 = 0

450 = 3 (x - 1) (x - 6)

t^{2} - 5 t - 5 = - 2 t

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

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