What are the solutions to the equation -16t^2+128t+320=0 ?

The solutions to the equation -16t^2+128t+320=0 are t=-2,t=10

Find the zeros of -16t^2+128t+320=0

The zeros of -16t^2+128t+320=0 are t=-2,t=10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 16 t^{2} + 128 t + 320 = 0

t = - 2, t = 10

Solution steps

- 16 t^{2} + 128 t + 320 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 16 ) \cdot 320}{2 ( - 16 )}

12 8^{2} - 4 (- 16) \cdot 320 = 192

t_{1, 2} = \frac{- 128 \pm 192}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 128 + 192}{2 ( - 16 )}, t_{2} = \frac{- 128 - 192}{2 ( - 16 )}

t = \frac{- 128 + 192}{2 ( - 16 )} : - 2

t = \frac{- 128 - 192}{2 ( - 16 )} : 10

The solutions to the quadratic equation are:

t = - 2, t = 10

- 6 (x - 3) > 4 x - 2

0 = - 4 p^{2} + 12000 p

3 y^{2} - 6 y - 45 = 0

6 x^{2} - 30 x + 24

(- \frac{4}{5})^{- 4}

What are the solutions to the equation -16t^2+128t+320=0 ?
The solutions to the equation -16t^2+128t+320=0 are t=-2,t=10
Find the zeros of -16t^2+128t+320=0
The zeros of -16t^2+128t+320=0 are t=-2,t=10