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

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

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

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

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

y = 0, y = - 1

Solution steps

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

For

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

the solutions are

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

Solve

4 y + 2 = 4 : y = 0

Solve

4 y + 2 = - 4 : y = - 1

The solutions to the quadratic equation are:

y = 0, y = - 1

\frac{7 ^{- 6}}{7 ^{- 2}}

x^{2} + 16 x + 64

0 = (x - \frac{1}{2})^{2} - \frac{81}{4}

13 = \frac{5}{2} (5 x + 4)

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