What are the solutions to the equation 3(y-4)^2=192 ?

The solutions to the equation 3(y-4)^2=192 are y=12,y=-4

Find the zeros of 3(y-4)^2=192

The zeros of 3(y-4)^2=192 are y=12,y=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 (y - 4)^{2} = 192

y = 12, y = - 4

Solution steps

3 (y - 4)^{2} = 192

Divide both sides by

3

(y - 4)^{2} = 64

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

y - 4 = 64 : y = 12

Solve

y - 4 = - 64 : y = - 4

The solutions to the quadratic equation are:

y = 12, y = - 4

- 3 x + 3 = x^{2} - x

5 x^{2} + 48 x + 27 = (5 x + 3)

6 (x - 3) - 6 (x - 4) = (x - 4) (x - 3)

80 \cdot 10 = \frac{v ^{2}}{2}

16.1 t^{2} - 200 t - 200 = 0

What are the solutions to the equation 3(y-4)^2=192 ?
The solutions to the equation 3(y-4)^2=192 are y=12,y=-4
Find the zeros of 3(y-4)^2=192
The zeros of 3(y-4)^2=192 are y=12,y=-4