What are the solutions to the equation (2^2}{25}+\frac{y^2)/9 =1 ?

The solutions to the equation (2^2}{25}+\frac{y^2)/9 =1 are y=(3sqrt(21))/5 ,y=-(3sqrt(21))/5

Find the zeros of (2^2}{25}+\frac{y^2)/9 =1

The zeros of (2^2}{25}+\frac{y^2)/9 =1 are y=(3sqrt(21))/5 ,y=-(3sqrt(21))/5

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\frac{2 ^{2}}{25} + \frac{y ^{2}}{9} = 1

y = \frac{3 21}{5}, y = - \frac{3 21}{5}

Solution steps

\frac{2 ^{2}}{25} + \frac{y ^{2}}{9} = 1

Move

\frac{2 ^{2}}{25}

to the right side

\frac{y ^{2}}{9} = 1 - \frac{4}{25}

Simplify

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

\frac{y ^{2}}{9} = \frac{21}{25}

Multiply both sides by

9

y^{2} = \frac{189}{25}

For

x^{2} = f (a)

the solutions are

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

y = \frac{189}{25}, y = - \frac{189}{25}

\frac{189}{25} = \frac{3 21}{5}

- \frac{189}{25} = - \frac{3 21}{5}

y = \frac{3 21}{5}, y = - \frac{3 21}{5}

g^{2} - 13 g - 48 = 0

x^{2} + 2.2 x + 1.21 = 0

2 0^{2} = 1 0^{2} + x^{2}

85.183 x^{2} - 174.9 x + 93.362 = 34

x^{2} - \frac{2}{3} x - \frac{1}{3} = 0

What are the solutions to the equation (2^2}{25}+\frac{y^2)/9 =1 ?
The solutions to the equation (2^2}{25}+\frac{y^2)/9 =1 are y=(3sqrt(21))/5 ,y=-(3sqrt(21))/5
Find the zeros of (2^2}{25}+\frac{y^2)/9 =1
The zeros of (2^2}{25}+\frac{y^2)/9 =1 are y=(3sqrt(21))/5 ,y=-(3sqrt(21))/5