What are the solutions to the equation 192=-16t^2+128 ?

The solutions to the equation 192=-16t^2+128 are t=2i,t=-2i

Find the zeros of 192=-16t^2+128

The zeros of 192=-16t^2+128 are t=2i,t=-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

192 = - 16 t^{2} + 128

t = 2 i, t = - 2 i

Solution steps

192 = - 16 t^{2} + 128

Switch sides

- 16 t^{2} + 128 = 192

Move

128

to the right side

- 16 t^{2} = 64

Divide both sides by

- 16

t^{2} = - 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

t = - 4, t = - - 4

- 4 = 2 i

- - 4 = - 2 i

t = 2 i, t = - 2 i

2 x^{2} + 16 x - 32

\frac{1}{- 1 + 2 i}

6 a^{2} - 23 a + 20

(\frac{1}{144})^{\frac{1}{2}}

∣ 2 a + 3 ∣ = 11

What are the solutions to the equation 192=-16t^2+128 ?
The solutions to the equation 192=-16t^2+128 are t=2i,t=-2i
Find the zeros of 192=-16t^2+128
The zeros of 192=-16t^2+128 are t=2i,t=-2i