What are the solutions to the equation 2((0-3)^2+(y+2)^2)=50 ?

The solutions to the equation 2((0-3)^2+(y+2)^2)=50 are y=2,y=-6

Find the zeros of 2((0-3)^2+(y+2)^2)=50

The zeros of 2((0-3)^2+(y+2)^2)=50 are y=2,y=-6

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2 ((0 - 3)^{2} + (y + 2)^{2}) = 50

y = 2, y = - 6

Solution steps

2 ((0 - 3)^{2} + (y + 2)^{2}) = 50

Divide both sides by

2

9 + (y + 2)^{2} = 25

Move

9

to the right side

(y + 2)^{2} = 16

For

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

the solutions are

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

Solve

y + 2 = 16 : y = 2

Solve

y + 2 = - 16 : y = - 6

The solutions to the quadratic equation are:

y = 2, y = - 6

10 x^{2} = 23 x + 5

12 x^{2} + 96 x = 0

3 n^{2} + 12 n - 15 = 0

2 g^{2} = 4 (17 + 8) + g^{2}

x = 0.25 x^{2} - 14.5 x + 200

What are the solutions to the equation 2((0-3)^2+(y+2)^2)=50 ?
The solutions to the equation 2((0-3)^2+(y+2)^2)=50 are y=2,y=-6
Find the zeros of 2((0-3)^2+(y+2)^2)=50
The zeros of 2((0-3)^2+(y+2)^2)=50 are y=2,y=-6