What are the solutions to the equation (0-4)^2+(y+2)^2=25 ?

The solutions to the equation (0-4)^2+(y+2)^2=25 are y=1,y=-5

Find the zeros of (0-4)^2+(y+2)^2=25

The zeros of (0-4)^2+(y+2)^2=25 are y=1,y=-5

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(0 - 4)^{2} + (y + 2)^{2} = 25

y = 1, y = - 5

Solution steps

(0 - 4)^{2} + (y + 2)^{2} = 25

Move

(0 - 4)^{2}

to the right side

(y + 2)^{2} = 9

For

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

the solutions are

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

Solve

y + 2 = 9 : y = 1

Solve

y + 2 = - 9 : y = - 5

The solutions to the quadratic equation are:

y = 1, y = - 5

30 (25 - t^{2}) = 400

(x - 8) (x - 3) (x + 2) > 0

3 - \frac{1}{2} x > 1

2 y^{2} - 6 y - 8 = 0

- 3 = \frac{h + 8}{2}

What are the solutions to the equation (0-4)^2+(y+2)^2=25 ?
The solutions to the equation (0-4)^2+(y+2)^2=25 are y=1,y=-5
Find the zeros of (0-4)^2+(y+2)^2=25
The zeros of (0-4)^2+(y+2)^2=25 are y=1,y=-5