What are the solutions to the equation (-7/3)^2+y^2=4 ?

The solutions to the equation (-7/3)^2+y^2=4 are y=(sqrt(13))/3 i,y=-(sqrt(13))/3 i

Find the zeros of (-7/3)^2+y^2=4

The zeros of (-7/3)^2+y^2=4 are y=(sqrt(13))/3 i,y=-(sqrt(13))/3 i

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(- \frac{7}{3})^{2} + y^{2} = 4

y = \frac{13}{3} i, y = - \frac{13}{3} i

Solution steps

(- \frac{7}{3})^{2} + y^{2} = 4

Expand

(- \frac{7}{3})^{2} + y^{2} : \frac{49}{9} + y^{2}

\frac{49}{9} + y^{2} = 4

Move

\frac{49}{9}

to the right side

y^{2} = 4 - \frac{49}{9}

Simplify

4 - \frac{49}{9} : - \frac{13}{9}

For

x^{2} = f (a)

the solutions are

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

y = - \frac{13}{9}, y = - - \frac{13}{9}

- \frac{13}{9} = \frac{13}{3} i

- - \frac{13}{9} = - \frac{13}{3} i

y = \frac{13}{3} i, y = - \frac{13}{3} i

x^{2} - 13 x + 48 = 0

y^{2} - y - 1 = 0

3 x^{2} - 12 x = 0

x, x^{2} + y^{2} - 4 x - 2 y + 9 = 0

x^{2} + 3 x - 4 = 0

What are the solutions to the equation (-7/3)^2+y^2=4 ?
The solutions to the equation (-7/3)^2+y^2=4 are y=(sqrt(13))/3 i,y=-(sqrt(13))/3 i
Find the zeros of (-7/3)^2+y^2=4
The zeros of (-7/3)^2+y^2=4 are y=(sqrt(13))/3 i,y=-(sqrt(13))/3 i