What are the solutions to the equation 272=-16t^2+32t ?

The solutions to the equation 272=-16t^2+32t are t=1-4i,t=1+4i

Find the zeros of 272=-16t^2+32t

The zeros of 272=-16t^2+32t are t=1-4i,t=1+4i

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272 = - 16 t^{2} + 32 t

t = 1 - 4 i, t = 1 + 4 i

Solution steps

272 = - 16 t^{2} + 32 t

Switch sides

- 16 t^{2} + 32 t = 272

Move

272

to the left side

- 16 t^{2} + 32 t - 272 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 16 ) ( - 272 )}{2 ( - 16 )}

Simplify

3 2^{2} - 4 (- 16) (- 272) : 128 i

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

Separate the solutions

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

t = \frac{- 32 + 128 i}{2 ( - 16 )} : 1 - 4 i

t = \frac{- 32 - 128 i}{2 ( - 16 )} : 1 + 4 i

The solutions to the quadratic equation are:

t = 1 - 4 i, t = 1 + 4 i

0 = 3 x^{2} + 13 x + 4

(x + 9)^{2} = - 5

- 3 x^{2} = 2 x

5 x + 2 x^{2} = 0

31500 = - 0.14 x^{2} + 420 x

What are the solutions to the equation 272=-16t^2+32t ?
The solutions to the equation 272=-16t^2+32t are t=1-4i,t=1+4i
Find the zeros of 272=-16t^2+32t
The zeros of 272=-16t^2+32t are t=1-4i,t=1+4i