What are the solutions to the equation (2/5)^2+y^2=1 ?

The solutions to the equation (2/5)^2+y^2=1 are y=(sqrt(21))/5 ,y=-(sqrt(21))/5

Find the zeros of (2/5)^2+y^2=1

The zeros of (2/5)^2+y^2=1 are y=(sqrt(21))/5 ,y=-(sqrt(21))/5

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(\frac{2}{5})^{2} + y^{2} = 1

y = \frac{21}{5}, y = - \frac{21}{5}

Decimal Notation

y = 0.91651 \dots, y = - 0.91651 \dots

Solution steps

(\frac{2}{5})^{2} + y^{2} = 1

Expand

(\frac{2}{5})^{2} + y^{2} : \frac{4}{25} + y^{2}

\frac{4}{25} + y^{2} = 1

Move

\frac{4}{25}

to the right side

y^{2} = 1 - \frac{4}{25}

Simplify

1 - \frac{4}{25} : \frac{21}{25}

For

x^{2} = f (a)

the solutions are

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

y = \frac{21}{25}, y = - \frac{21}{25}

\frac{21}{25} = \frac{21}{5}

- \frac{21}{25} = - \frac{21}{5}

y = \frac{21}{5}, y = - \frac{21}{5}

x, x^{2} - 2 x - 8 = 0

s^{2} + 8 s + 8 = 0

15 x^{2} + 34 x - 77 = 0

3 x^{1} + 6 x^{2} = 9, - 2 x^{1} + 3 x^{2} = 4

\frac{x ^{2} - x}{2} - \frac{( x - 4 ) ^{2}}{3} = 0

What are the solutions to the equation (2/5)^2+y^2=1 ?
The solutions to the equation (2/5)^2+y^2=1 are y=(sqrt(21))/5 ,y=-(sqrt(21))/5
Find the zeros of (2/5)^2+y^2=1
The zeros of (2/5)^2+y^2=1 are y=(sqrt(21))/5 ,y=-(sqrt(21))/5