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

The solutions to the equation (y+1)^2=4 are y=1,y=-3

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

The zeros of (y+1)^2=4 are y=1,y=-3

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(y + 1)^{2} = 4

y = 1, y = - 3

Solution steps

(y + 1)^{2} = 4

For

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

the solutions are

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

Solve

y + 1 = 4 : y = 1

Solve

y + 1 = - 4 : y = - 3

The solutions to the quadratic equation are:

y = 1, y = - 3

- 16 t^{2} + 70 t + 5 = 50

\frac{3 x - 2 x}{1 + x} \leq 1

- 4 v \leq - 24 or 2 v - 3 < - 15

255

6 n^{2} - 30 n + 42

What are the solutions to the equation (y+1)^2=4 ?
The solutions to the equation (y+1)^2=4 are y=1,y=-3
Find the zeros of (y+1)^2=4
The zeros of (y+1)^2=4 are y=1,y=-3