What are the solutions to the equation -16t^2+560t=0 ?

The solutions to the equation -16t^2+560t=0 are t=0,t=35

Find the zeros of -16t^2+560t=0

The zeros of -16t^2+560t=0 are t=0,t=35

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- 16 t^{2} + 560 t = 0

t = 0, t = 35

Solution steps

- 16 t^{2} + 560 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 560 \pm 56 0 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

56 0^{2} - 4 (- 16) \cdot 0 = 560

t_{1, 2} = \frac{- 560 \pm 560}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 560 + 560}{2 ( - 16 )}, t_{2} = \frac{- 560 - 560}{2 ( - 16 )}

t = \frac{- 560 + 560}{2 ( - 16 )} : 0

t = \frac{- 560 - 560}{2 ( - 16 )} : 35

The solutions to the quadratic equation are:

t = 0, t = 35

16 x^{2} - 120 x + 180 = 0

x^{2} - 15 x + 26 = - 3 x - 6

8 x^{2} + 24 x - 20 = 0

2 x^{2} - 5 x - 7 = 0

(- x - 1)^{2} = 0

What are the solutions to the equation -16t^2+560t=0 ?
The solutions to the equation -16t^2+560t=0 are t=0,t=35
Find the zeros of -16t^2+560t=0
The zeros of -16t^2+560t=0 are t=0,t=35