What are the solutions to the equation 3y^2+2=0 ?

The solutions to the equation 3y^2+2=0 are y=sqrt(2/3)i,y=-sqrt(2/3)i

Find the zeros of 3y^2+2=0

The zeros of 3y^2+2=0 are y=sqrt(2/3)i,y=-sqrt(2/3)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 y^{2} + 2 = 0

y = \frac{2}{3} i, y = - \frac{2}{3} i

Solution steps

3 y^{2} + 2 = 0

Move

2

to the right side

3 y^{2} = - 2

Divide both sides by

3

y^{2} = - \frac{2}{3}

For

x^{2} = f (a)

the solutions are

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

y = - \frac{2}{3}, y = - - \frac{2}{3}

- \frac{2}{3} = \frac{2}{3} i

- - \frac{2}{3} = - \frac{2}{3} i

y = \frac{2}{3} i, y = - \frac{2}{3} i

4 x^{2} + 12 x - 44 = 0

0.51 = \frac{12.96 + 28.09 - c ^{2}}{38.16}

2 p^{2} - 6 p + 7 = 0

- 4 x^{2} = - 64

w^{2} - 5 w - 14 = 0

What are the solutions to the equation 3y^2+2=0 ?
The solutions to the equation 3y^2+2=0 are y=sqrt(2/3)i,y=-sqrt(2/3)i
Find the zeros of 3y^2+2=0
The zeros of 3y^2+2=0 are y=sqrt(2/3)i,y=-sqrt(2/3)i