What are the solutions to the equation (0-3)^2+(y+5)^2=16 ?

The solutions to the equation (0-3)^2+(y+5)^2=16 are y=sqrt(7)-5,y=-sqrt(7)-5

Find the zeros of (0-3)^2+(y+5)^2=16

The zeros of (0-3)^2+(y+5)^2=16 are y=sqrt(7)-5,y=-sqrt(7)-5

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(0 - 3)^{2} + (y + 5)^{2} = 16

y = 7 - 5, y = - 7 - 5

Decimal Notation

y = - 2.35424 \dots, y = - 7.64575 \dots

Solution steps

(0 - 3)^{2} + (y + 5)^{2} = 16

Move

(0 - 3)^{2}

to the right side

(y + 5)^{2} = 7

For

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

the solutions are

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

Solve

y + 5 = 7 : y = 7 - 5

Solve

y + 5 = - 7 : y = - 7 - 5

The solutions to the quadratic equation are:

y = 7 - 5, y = - 7 - 5

∣ 2 x - 3 ∣ \leq 5

x^{2} + a^{2} + 2 a x

13 - 6 n = - 77

9 x^{2} + 6 x + 12 x + 8

x (- y) (- z)

What are the solutions to the equation (0-3)^2+(y+5)^2=16 ?
The solutions to the equation (0-3)^2+(y+5)^2=16 are y=sqrt(7)-5,y=-sqrt(7)-5
Find the zeros of (0-3)^2+(y+5)^2=16
The zeros of (0-3)^2+(y+5)^2=16 are y=sqrt(7)-5,y=-sqrt(7)-5