What are the solutions to the equation (-(sqrt(15))/4)^2+y^2=1 ?

The solutions to the equation (-(sqrt(15))/4)^2+y^2=1 are y= 1/4 ,y=-1/4

Find the zeros of (-(sqrt(15))/4)^2+y^2=1

The zeros of (-(sqrt(15))/4)^2+y^2=1 are y= 1/4 ,y=-1/4

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(- \frac{15}{4})^{2} + y^{2} = 1

y = \frac{1}{4}, y = - \frac{1}{4}

Decimal Notation

y = 0.25, y = - 0.25

Solution steps

(- \frac{15}{4})^{2} + y^{2} = 1

Expand

(- \frac{15}{4})^{2} + y^{2} : \frac{15}{16} + y^{2}

\frac{15}{16} + y^{2} = 1

Move

\frac{15}{16}

to the right side

y^{2} = 1 - \frac{15}{16}

Simplify

1 - \frac{15}{16} : \frac{1}{16}

For

x^{2} = f (a)

the solutions are

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

y = \frac{1}{16}, y = - \frac{1}{16}

\frac{1}{16} = \frac{1}{4}

- \frac{1}{16} = - \frac{1}{4}

y = \frac{1}{4}, y = - \frac{1}{4}

4 x^{2} - 6 x - 11 = 0

x, (a x + 4 y)^{2} = 25 x^{2} - 40 x y + 16 y^{2}

x^{2} + (7 - x) 2 = 25

x^{2} + x - 72 = 0

1.6 x^{2} + 2 x + 2.5 = 0

What are the solutions to the equation (-(sqrt(15))/4)^2+y^2=1 ?
The solutions to the equation (-(sqrt(15))/4)^2+y^2=1 are y= 1/4 ,y=-1/4
Find the zeros of (-(sqrt(15))/4)^2+y^2=1
The zeros of (-(sqrt(15))/4)^2+y^2=1 are y= 1/4 ,y=-1/4