What are the solutions to the equation solvefor y,x^2+(y+3/4)^2=1 ?

The solutions to the equation solvefor y,x^2+(y+3/4)^2=1 are y=sqrt(1-x^2)-3/4 ,y=-sqrt(1-x^2)-3/4

Find the zeros of solvefor y,x^2+(y+3/4)^2=1

The zeros of solvefor y,x^2+(y+3/4)^2=1 are y=sqrt(1-x^2)-3/4 ,y=-sqrt(1-x^2)-3/4

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solve for

y, x^{2} + (y + \frac{3}{4})^{2} = 1

y = 1 - x^{2} - \frac{3}{4}, y = - 1 - x^{2} - \frac{3}{4}

Solution steps

x^{2} + (y + \frac{3}{4})^{2} = 1

Move

x^{2}

to the right side

(y + \frac{3}{4})^{2} = 1 - x^{2}

For

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

the solutions are

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

Solve

y + \frac{3}{4} = 1 - x^{2} : y = 1 - x^{2} - \frac{3}{4}

Solve

y + \frac{3}{4} = - 1 - x^{2} : y = - 1 - x^{2} - \frac{3}{4}

The solutions to the quadratic equation are:

y = 1 - x^{2} - \frac{3}{4}, y = - 1 - x^{2} - \frac{3}{4}

4 p^{2} + 23 = 7

x^{2} = 0.1

b, N b^{2} - r = 3 T

x^{2} = 17 x

6 x^{2} + 6 x + 36 = 0

What are the solutions to the equation solvefor y,x^2+(y+3/4)^2=1 ?
The solutions to the equation solvefor y,x^2+(y+3/4)^2=1 are y=sqrt(1-x^2)-3/4 ,y=-sqrt(1-x^2)-3/4
Find the zeros of solvefor y,x^2+(y+3/4)^2=1
The zeros of solvefor y,x^2+(y+3/4)^2=1 are y=sqrt(1-x^2)-3/4 ,y=-sqrt(1-x^2)-3/4