What are the solutions to the equation (3y+2)^2+10=-18 ?

The solutions to the equation (3y+2)^2+10=-18 are y=-2/3+i(2sqrt(7))/3 ,y=-2/3-i(2sqrt(7))/3

Find the zeros of (3y+2)^2+10=-18

The zeros of (3y+2)^2+10=-18 are y=-2/3+i(2sqrt(7))/3 ,y=-2/3-i(2sqrt(7))/3

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(3 y + 2)^{2} + 10 = - 18

y = - \frac{2}{3} + i \frac{2 7}{3}, y = - \frac{2}{3} - i \frac{2 7}{3}

Solution steps

(3 y + 2)^{2} + 10 = - 18

Move

10

to the right side

(3 y + 2)^{2} = - 28

For

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

the solutions are

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

Solve

3 y + 2 = - 28 : y = - \frac{2}{3} + i \frac{2 7}{3}

Solve

3 y + 2 = - - 28 : y = - \frac{2}{3} - i \frac{2 7}{3}

The solutions to the quadratic equation are:

y = - \frac{2}{3} + i \frac{2 7}{3}, y = - \frac{2}{3} - i \frac{2 7}{3}

1 = 2

\frac{k ^{2} + 9 k - 10}{34} \cdot \frac{2}{k ^{2} - 1}

3 x + 1 = x + 8

5 x^{2} - 6 x + 2 = 0

x^{\frac{3}{2}}

What are the solutions to the equation (3y+2)^2+10=-18 ?
The solutions to the equation (3y+2)^2+10=-18 are y=-2/3+i(2sqrt(7))/3 ,y=-2/3-i(2sqrt(7))/3
Find the zeros of (3y+2)^2+10=-18
The zeros of (3y+2)^2+10=-18 are y=-2/3+i(2sqrt(7))/3 ,y=-2/3-i(2sqrt(7))/3