What are the solutions to the equation (0+4)^2+(y+5)^2=81 ?

The solutions to the equation (0+4)^2+(y+5)^2=81 are y=sqrt(65)-5,y=-sqrt(65)-5

Find the zeros of (0+4)^2+(y+5)^2=81

The zeros of (0+4)^2+(y+5)^2=81 are y=sqrt(65)-5,y=-sqrt(65)-5

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(0 + 4)^{2} + (y + 5)^{2} = 81

y = 65 - 5, y = - 65 - 5

Solution steps

(0 + 4)^{2} + (y + 5)^{2} = 81

Move

(0 + 4)^{2}

to the right side

(y + 5)^{2} = 65

For

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

the solutions are

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

Solve

y + 5 = 65 : y = 65 - 5

Solve

y + 5 = - 65 : y = - 65 - 5

The solutions to the quadratic equation are:

y = 65 - 5, y = - 65 - 5

5 - \frac{x}{12} = 8

\frac{2}{3} (\frac{1}{6} x - \frac{3}{4}) = - \frac{17}{30}

(\frac{3 x ^{4}}{2 y ^{3}})^{3}

25 6^{\frac{3}{8}}

0.16 y + 0.06 (y + 5000) = 960

What are the solutions to the equation (0+4)^2+(y+5)^2=81 ?
The solutions to the equation (0+4)^2+(y+5)^2=81 are y=sqrt(65)-5,y=-sqrt(65)-5
Find the zeros of (0+4)^2+(y+5)^2=81
The zeros of (0+4)^2+(y+5)^2=81 are y=sqrt(65)-5,y=-sqrt(65)-5