What are the solutions to the equation 192=-16t^2+112t ?

The solutions to the equation 192=-16t^2+112t are t=3,t=4

Find the zeros of 192=-16t^2+112t

The zeros of 192=-16t^2+112t are t=3,t=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

192 = - 16 t^{2} + 112 t

t = 3, t = 4

Solution steps

192 = - 16 t^{2} + 112 t

Switch sides

- 16 t^{2} + 112 t = 192

Move

192

to the left side

- 16 t^{2} + 112 t - 192 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 112 \pm 11 2 ^{2} - 4 ( - 16 ) ( - 192 )}{2 ( - 16 )}

11 2^{2} - 4 (- 16) (- 192) = 16

t_{1, 2} = \frac{- 112 \pm 16}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 112 + 16}{2 ( - 16 )}, t_{2} = \frac{- 112 - 16}{2 ( - 16 )}

t = \frac{- 112 + 16}{2 ( - 16 )} : 3

t = \frac{- 112 - 16}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = 3, t = 4

- 2 x^{2} = 8 x

(\frac{5}{13})^{2} + y^{2} = 1

y, (3 yz + 8) (4 yz + 4) = 0

361 x^{2} - 380 x + 100 = 0

4 x^{2} - 60 = 4

What are the solutions to the equation 192=-16t^2+112t ?
The solutions to the equation 192=-16t^2+112t are t=3,t=4
Find the zeros of 192=-16t^2+112t
The zeros of 192=-16t^2+112t are t=3,t=4