What are the solutions to the equation (0+2)^2+(y-5)^2=25 ?

The solutions to the equation (0+2)^2+(y-5)^2=25 are y=sqrt(21)+5,y=-sqrt(21)+5

Find the zeros of (0+2)^2+(y-5)^2=25

The zeros of (0+2)^2+(y-5)^2=25 are y=sqrt(21)+5,y=-sqrt(21)+5

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(0 + 2)^{2} + (y - 5)^{2} = 25

y = 21 + 5, y = - 21 + 5

Decimal Notation

y = 9.58257 \dots, y = 0.41742 \dots

Solution steps

(0 + 2)^{2} + (y - 5)^{2} = 25

Move

(0 + 2)^{2}

to the right side

(y - 5)^{2} = 21

For

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

the solutions are

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

Solve

y - 5 = 21 : y = 21 + 5

Solve

y - 5 = - 21 : y = - 21 + 5

The solutions to the quadratic equation are:

y = 21 + 5, y = - 21 + 5

(a - 7)^{2} + 5 = 55

4 x (x + 8) = 0

p^{2} - 16 p + 16 = - 8 p

5 x^{2} - 6 x - 2 = x (3 x - 3)

2 x^{2} + 10 = 12 x

What are the solutions to the equation (0+2)^2+(y-5)^2=25 ?
The solutions to the equation (0+2)^2+(y-5)^2=25 are y=sqrt(21)+5,y=-sqrt(21)+5
Find the zeros of (0+2)^2+(y-5)^2=25
The zeros of (0+2)^2+(y-5)^2=25 are y=sqrt(21)+5,y=-sqrt(21)+5