What are the solutions to the equation (-3/4)^2+y^2=1 ?

The solutions to the equation (-3/4)^2+y^2=1 are y=(sqrt(7))/4 ,y=-(sqrt(7))/4

Find the zeros of (-3/4)^2+y^2=1

The zeros of (-3/4)^2+y^2=1 are y=(sqrt(7))/4 ,y=-(sqrt(7))/4

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(- \frac{3}{4})^{2} + y^{2} = 1

y = \frac{7}{4}, y = - \frac{7}{4}

Decimal Notation

y = 0.66143 \dots, y = - 0.66143 \dots

Solution steps

(- \frac{3}{4})^{2} + y^{2} = 1

Expand

(- \frac{3}{4})^{2} + y^{2} : \frac{9}{16} + y^{2}

\frac{9}{16} + y^{2} = 1

Move

\frac{9}{16}

to the right side

y^{2} = 1 - \frac{9}{16}

Simplify

1 - \frac{9}{16} : \frac{7}{16}

For

x^{2} = f (a)

the solutions are

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

y = \frac{7}{16}, y = - \frac{7}{16}

\frac{7}{16} = \frac{7}{4}

- \frac{7}{16} = - \frac{7}{4}

y = \frac{7}{4}, y = - \frac{7}{4}

- 3 x^{2} - 7 x + 2 = 6

0 = 4 x^{2} - 8 x - 5

D, D^{2} y - Dy - 6 y = e^{2 x}

x, 3 x^{2} + 4 y^{2} - 6 x - 24 y + 39 = 0

x, x^{2} + 2 x + y^{2} - 6 y = c

What are the solutions to the equation (-3/4)^2+y^2=1 ?
The solutions to the equation (-3/4)^2+y^2=1 are y=(sqrt(7))/4 ,y=-(sqrt(7))/4
Find the zeros of (-3/4)^2+y^2=1
The zeros of (-3/4)^2+y^2=1 are y=(sqrt(7))/4 ,y=-(sqrt(7))/4