What are the solutions to the equation (y+1)^2-9=0 ?

The solutions to the equation (y+1)^2-9=0 are y=2,y=-4

Find the zeros of (y+1)^2-9=0

The zeros of (y+1)^2-9=0 are y=2,y=-4

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(y + 1)^{2} - 9 = 0

y = 2, y = - 4

Solution steps

(y + 1)^{2} - 9 = 0

Move

9

to the right side

(y + 1)^{2} = 9

For

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

the solutions are

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

Solve

y + 1 = 9 : y = 2

Solve

y + 1 = - 9 : y = - 4

The solutions to the quadratic equation are:

y = 2, y = - 4

x^{2} - 12 x - 26 = 0

x^{2} - 1 = lo g_{3} (27)

4 x^{2} + 18 x + 18 = 0

x^{2} + 18 x + 81 = 49

x^{2} - 1 = 6 x

What are the solutions to the equation (y+1)^2-9=0 ?
The solutions to the equation (y+1)^2-9=0 are y=2,y=-4
Find the zeros of (y+1)^2-9=0
The zeros of (y+1)^2-9=0 are y=2,y=-4