What are the solutions to the equation (0-1)^2+(y+8)^2=4 ?

The solutions to the equation (0-1)^2+(y+8)^2=4 are y=sqrt(3)-8,y=-sqrt(3)-8

Find the zeros of (0-1)^2+(y+8)^2=4

The zeros of (0-1)^2+(y+8)^2=4 are y=sqrt(3)-8,y=-sqrt(3)-8

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(0 - 1)^{2} + (y + 8)^{2} = 4

y = 3 - 8, y = - 3 - 8

Decimal Notation

y = - 6.26794 \dots, y = - 9.73205 \dots

Solution steps

(0 - 1)^{2} + (y + 8)^{2} = 4

Move

(0 - 1)^{2}

to the right side

(y + 8)^{2} = 3

For

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

the solutions are

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

Solve

y + 8 = 3 : y = 3 - 8

Solve

y + 8 = - 3 : y = - 3 - 8

The solutions to the quadratic equation are:

y = 3 - 8, y = - 3 - 8

\frac{x ^{2} - 4 x + 3}{x ^{2} - 6 x + 8} > 0

x^{2} \leq 2 x

- 3 x^{2} + 10 x - 5 = 0

\frac{13}{9} x^{2} - \frac{40}{3} x - 900 = 0

0 = x^{2} - 4

What are the solutions to the equation (0-1)^2+(y+8)^2=4 ?
The solutions to the equation (0-1)^2+(y+8)^2=4 are y=sqrt(3)-8,y=-sqrt(3)-8
Find the zeros of (0-1)^2+(y+8)^2=4
The zeros of (0-1)^2+(y+8)^2=4 are y=sqrt(3)-8,y=-sqrt(3)-8