What are the solutions to the equation (4y+4)^2=-16 ?

The solutions to the equation (4y+4)^2=-16 are y=i-1,y=-1-i

Find the zeros of (4y+4)^2=-16

The zeros of (4y+4)^2=-16 are y=i-1,y=-1-i

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(4 y + 4)^{2} = - 16

y = i - 1, y = - 1 - i

Solution steps

(4 y + 4)^{2} = - 16

For

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

the solutions are

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

Solve

4 y + 4 = - 16 : y = i - 1

Solve

4 y + 4 = - - 16 : y = - 1 - i

The solutions to the quadratic equation are:

y = i - 1, y = - 1 - i

3 x^{2} - 96 x + 92 = 0

x (x - 12) + 32 = 0

x, x^{2} - x y + 5 y = 0

x, s (- 1) = x^{2} - 4 x - 21

x^{2} + 10 x + 37 = 0

What are the solutions to the equation (4y+4)^2=-16 ?
The solutions to the equation (4y+4)^2=-16 are y=i-1,y=-1-i
Find the zeros of (4y+4)^2=-16
The zeros of (4y+4)^2=-16 are y=i-1,y=-1-i