What are the solutions to the equation y^2=2*1/16 (3+1) ?

The solutions to the equation y^2=2*1/16 (3+1) are y= 1/(sqrt(2)),y=-1/(sqrt(2))

Find the zeros of y^2=2*1/16 (3+1)

The zeros of y^2=2*1/16 (3+1) are y= 1/(sqrt(2)),y=-1/(sqrt(2))

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y^{2} = 2 \cdot \frac{1}{16} (3 + 1)

y = \frac{1}{2}, y = - \frac{1}{2}

Decimal Notation

y = 0.70710 \dots, y = - 0.70710 \dots

Solution steps

y^{2} = 2 \cdot \frac{1}{16} (3 + 1)

Simplify

2 \cdot \frac{1}{16} (3 + 1) : \frac{1}{2}

For

x^{2} = f (a)

the solutions are

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

y = \frac{1}{2}, y = - \frac{1}{2}

\frac{1}{2} = \frac{1}{2}

- \frac{1}{2} = - \frac{1}{2}

y = \frac{1}{2}, y = - \frac{1}{2}

3 y^{2} - 14 y = 5

3 x^{2} - 8 x - 12 = 0

x^{2} + 30 = 24 x - 114

0 = - 407 x^{2} + 220275 x - 19824750

4 x^{2} + 28 x = 72

What are the solutions to the equation y^2=2*1/16 (3+1) ?
The solutions to the equation y^2=2*1/16 (3+1) are y= 1/(sqrt(2)),y=-1/(sqrt(2))
Find the zeros of y^2=2*1/16 (3+1)
The zeros of y^2=2*1/16 (3+1) are y= 1/(sqrt(2)),y=-1/(sqrt(2))