What are the solutions to the equation 18=(-6+3)^2+(2-y)^2 ?

The solutions to the equation 18=(-6+3)^2+(2-y)^2 are y=-1,y=5

Find the zeros of 18=(-6+3)^2+(2-y)^2

The zeros of 18=(-6+3)^2+(2-y)^2 are y=-1,y=5

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18 = (- 6 + 3)^{2} + (2 - y)^{2}

y = - 1, y = 5

Solution steps

18 = (- 6 + 3)^{2} + (2 - y)^{2}

Switch sides

(- 6 + 3)^{2} + (2 - y)^{2} = 18

Move

(- 6 + 3)^{2}

to the right side

(2 - y)^{2} = 9

For

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

the solutions are

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

Solve

2 - y = 9 : y = - 1

Solve

2 - y = - 9 : y = 5

The solutions to the quadratic equation are:

y = - 1, y = 5

(3 n - 2) = \frac{n ( 3 n - 1 )}{2}

u^{2} - 2 u - 2 = 0

x^{2} - 2 x - c = 0

(2 x - 1)^{2} = (x + 4)^{2}

x^{2} - 4.649160568 x + 3.125 = 0

What are the solutions to the equation 18=(-6+3)^2+(2-y)^2 ?
The solutions to the equation 18=(-6+3)^2+(2-y)^2 are y=-1,y=5
Find the zeros of 18=(-6+3)^2+(2-y)^2
The zeros of 18=(-6+3)^2+(2-y)^2 are y=-1,y=5