What are the solutions to the equation 1^2+y^2=2^2 ?

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

Find the zeros of 1^2+y^2=2^2

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

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1^{2} + y^{2} = 2^{2}

y = 3, y = - 3

Decimal Notation

y = 1.73205 \dots, y = - 1.73205 \dots

Solution steps

1^{2} + y^{2} = 2^{2}

Move

1^{2}

to the right side

y^{2} = 3

For

x^{2} = f (a)

the solutions are

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

y = 3, y = - 3

x^{2} + (2 x + 3)^{2} = 24

2 0^{2} + 7^{2} = x^{2}

t, e = e_{0} + v_{0} \cdot t + \frac{1}{2} \cdot a t^{2}

y, x^{2} + (y - 1)^{2} = 4

x, 2 x^{2} - 5 x y - 3 y^{2} = 0

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