What are the solutions to the equation 0=-16t^2+16t+320 ?

The solutions to the equation 0=-16t^2+16t+320 are t=-4,t=5

Find the zeros of 0=-16t^2+16t+320

The zeros of 0=-16t^2+16t+320 are t=-4,t=5

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0 = - 16 t^{2} + 16 t + 320

t = - 4, t = 5

Solution steps

0 = - 16 t^{2} + 16 t + 320

Switch sides

- 16 t^{2} + 16 t + 320 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) \cdot 320}{2 ( - 16 )}

1 6^{2} - 4 (- 16) \cdot 320 = 144

t_{1, 2} = \frac{- 16 \pm 144}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 16 + 144}{2 ( - 16 )}, t_{2} = \frac{- 16 - 144}{2 ( - 16 )}

t = \frac{- 16 + 144}{2 ( - 16 )} : - 4

t = \frac{- 16 - 144}{2 ( - 16 )} : 5

The solutions to the quadratic equation are:

t = - 4, t = 5

8 - p^{2} = p^{2}

x - 1 = x^{2} - 1

x^{2} + 160 = 0

s^{2} 5 + s + 5 = 0

0 = 3 x^{2} + 18 x + 27

What are the solutions to the equation 0=-16t^2+16t+320 ?
The solutions to the equation 0=-16t^2+16t+320 are t=-4,t=5
Find the zeros of 0=-16t^2+16t+320
The zeros of 0=-16t^2+16t+320 are t=-4,t=5