What are the solutions to the equation (-(sqrt(3))/2)^2+y^2=1 ?

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

Find the zeros of (-(sqrt(3))/2)^2+y^2=1

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

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

y = \frac{1}{2}, y = - \frac{1}{2}

Decimal Notation

y = 0.5, y = - 0.5

Solution steps

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

Expand

(- \frac{3}{2})^{2} + y^{2} : \frac{3}{4} + y^{2}

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

Move

\frac{3}{4}

to the right side

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

Simplify

1 - \frac{3}{4} : \frac{1}{4}

For

x^{2} = f (a)

the solutions are

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

y = \frac{1}{4}, y = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

y = \frac{1}{2}, y = - \frac{1}{2}

4 x^{2} - 24 x + 100 = 0

56 (1000) - 0.02 (1000)^{2}

\frac{x}{5} - \frac{2 x}{15} \geq \frac{x + 4}{3}

\frac{2}{5} x - 4 = - \frac{1}{5} x + 8

10 y^{2} - 21 y - 10

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