What are the solutions to the equation (0-2)^2+(y+3)^2=9 ?

The solutions to the equation (0-2)^2+(y+3)^2=9 are y=sqrt(5)-3,y=-sqrt(5)-3

Find the zeros of (0-2)^2+(y+3)^2=9

The zeros of (0-2)^2+(y+3)^2=9 are y=sqrt(5)-3,y=-sqrt(5)-3

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(0 - 2)^{2} + (y + 3)^{2} = 9

y = 5 - 3, y = - 5 - 3

Decimal Notation

y = - 0.76393 \dots, y = - 5.23606 \dots

Solution steps

(0 - 2)^{2} + (y + 3)^{2} = 9

Move

(0 - 2)^{2}

to the right side

(y + 3)^{2} = 5

For

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

the solutions are

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

Solve

y + 3 = 5 : y = 5 - 3

Solve

y + 3 = - 5 : y = - 5 - 3

The solutions to the quadratic equation are:

y = 5 - 3, y = - 5 - 3

2 x^{2} + 16 x + 4 = 0

a^{2} + 7 a + 2 = 0

\frac{2}{3} x^{2} + \frac{1}{4} x = 0

0 = - 16 t^{2} + 64 t + 336

\frac{1}{2} x + \frac{1}{2} = x^{2} - 1

What are the solutions to the equation (0-2)^2+(y+3)^2=9 ?
The solutions to the equation (0-2)^2+(y+3)^2=9 are y=sqrt(5)-3,y=-sqrt(5)-3
Find the zeros of (0-2)^2+(y+3)^2=9
The zeros of (0-2)^2+(y+3)^2=9 are y=sqrt(5)-3,y=-sqrt(5)-3