What are the solutions to the equation (0-3)^2+(y-4)^2=36 ?

The solutions to the equation (0-3)^2+(y-4)^2=36 are y=3sqrt(3)+4,y=-3sqrt(3)+4

Find the zeros of (0-3)^2+(y-4)^2=36

The zeros of (0-3)^2+(y-4)^2=36 are y=3sqrt(3)+4,y=-3sqrt(3)+4

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(0 - 3)^{2} + (y - 4)^{2} = 36

y = 33 + 4, y = - 33 + 4

Decimal Notation

y = 9.19615 \dots, y = - 1.19615 \dots

Solution steps

(0 - 3)^{2} + (y - 4)^{2} = 36

Move

(0 - 3)^{2}

to the right side

(y - 4)^{2} = 27

For

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

the solutions are

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

Solve

y - 4 = 27 : y = 33 + 4

Solve

y - 4 = - 27 : y = - 33 + 4

The solutions to the quadratic equation are:

y = 33 + 4, y = - 33 + 4

(\frac{1}{4})^{- \frac{3}{2}}

4 k^{2} + 4 k - 2

1 \leq - 1 - f

\frac{6 + 2 i}{3 - i}

\frac{2 ^{8}}{2 ^{3}}

What are the solutions to the equation (0-3)^2+(y-4)^2=36 ?
The solutions to the equation (0-3)^2+(y-4)^2=36 are y=3sqrt(3)+4,y=-3sqrt(3)+4
Find the zeros of (0-3)^2+(y-4)^2=36
The zeros of (0-3)^2+(y-4)^2=36 are y=3sqrt(3)+4,y=-3sqrt(3)+4