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

The solutions to the equation (0-1)^2+(y-(-2))^2=2^2 are y=sqrt(3)-2,y=-sqrt(3)-2

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

The zeros of (0-1)^2+(y-(-2))^2=2^2 are y=sqrt(3)-2,y=-sqrt(3)-2

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

y = 3 - 2, y = - 3 - 2

Solution steps

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

Move

(0 - 1)^{2}

to the right side

(y - (- 2))^{2} = 3

For

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

the solutions are

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

Solve

y - (- 2) = 3 : y = 3 - 2

Solve

y - (- 2) = - 3 : y = - 3 - 2

The solutions to the quadratic equation are:

y = 3 - 2, y = - 3 - 2

\frac{1}{2} + \frac{3}{1}

20 v^{7} w^{9} - 15 v w^{6} y^{2}

\frac{1}{e ^{0}}

x^{2} - 3 x = 8

\frac{36 c ^{5} d ^{4}}{96 c d ^{8}}

What are the solutions to the equation (0-1)^2+(y-(-2))^2=2^2 ?
The solutions to the equation (0-1)^2+(y-(-2))^2=2^2 are y=sqrt(3)-2,y=-sqrt(3)-2
Find the zeros of (0-1)^2+(y-(-2))^2=2^2
The zeros of (0-1)^2+(y-(-2))^2=2^2 are y=sqrt(3)-2,y=-sqrt(3)-2