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

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

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

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

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

y = 3, y = - 3

Decimal Notation

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

Solution steps

- (2)^{2} + y^{2} = 1

Move

2^{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

(2 x^{3})^{5}

12 x + 18 y + 26

x (7 x - 3) = 10

(x + 2)^{2} - (x + 3)^{2}

275 = 5.5 t - 110

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