What are the solutions to the equation (7y+1)^2+4=0 ?

The solutions to the equation (7y+1)^2+4=0 are y=-1/7+i 2/7 ,y=-1/7-i 2/7

Find the zeros of (7y+1)^2+4=0

The zeros of (7y+1)^2+4=0 are y=-1/7+i 2/7 ,y=-1/7-i 2/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(7 y + 1)^{2} + 4 = 0

y = - \frac{1}{7} + i \frac{2}{7}, y = - \frac{1}{7} - i \frac{2}{7}

Solution steps

(7 y + 1)^{2} + 4 = 0

Move

4

to the right side

(7 y + 1)^{2} = - 4

For

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

the solutions are

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

Solve

7 y + 1 = - 4 : y = - \frac{1}{7} + i \frac{2}{7}

Solve

7 y + 1 = - - 4 : y = - \frac{1}{7} - i \frac{2}{7}

The solutions to the quadratic equation are:

y = - \frac{1}{7} + i \frac{2}{7}, y = - \frac{1}{7} - i \frac{2}{7}

4 x^{2} + 19 x - 5 = 0

0 - 6

\frac{a ^{3} b ^{- 2}}{a b ^{- 4}}

- \frac{5}{8} \cdot 3

10 = t - 8

What are the solutions to the equation (7y+1)^2+4=0 ?
The solutions to the equation (7y+1)^2+4=0 are y=-1/7+i 2/7 ,y=-1/7-i 2/7
Find the zeros of (7y+1)^2+4=0
The zeros of (7y+1)^2+4=0 are y=-1/7+i 2/7 ,y=-1/7-i 2/7