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

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

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

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

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

r = 1, r = - 1

Solution steps

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

Switch sides

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

Simplify

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

For

x^{2} = f (a)

the solutions are

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

r = 1, r = - 1

1 = 1

- 1 = - 1

r = 1, r = - 1

3 x^{2} - 6 x + 2 = 0

2 x^{2} - 8 = 41

3 x^{2} - x - 48 = 0

- 0.6 t^{2} + 48 t = 0

2 z^{2} + 3 z - 6 = 0

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