What are the solutions to the equation x^2+(-3/6)^2=1 ?

The solutions to the equation x^2+(-3/6)^2=1 are x=(sqrt(3))/2 ,x=-(sqrt(3))/2

Find the zeros of x^2+(-3/6)^2=1

The zeros of x^2+(-3/6)^2=1 are x=(sqrt(3))/2 ,x=-(sqrt(3))/2

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x^{2} + (- \frac{3}{6})^{2} = 1

x = \frac{3}{2}, x = - \frac{3}{2}

Decimal Notation

x = 0.86602 \dots, x = - 0.86602 \dots

Solution steps

x^{2} + (- \frac{3}{6})^{2} = 1

Expand

x^{2} + (- \frac{3}{6})^{2} : x^{2} + \frac{1}{4}

x^{2} + \frac{1}{4} = 1

Move

\frac{1}{4}

to the right side

x^{2} = 1 - \frac{1}{4}

Simplify

1 - \frac{1}{4} : \frac{3}{4}

For

x^{2} = f (a)

the solutions are

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

x = \frac{3}{4}, x = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

x = \frac{3}{2}, x = - \frac{3}{2}

5 x^{2} = 3 x + 8

t^{2} + \frac{11}{3} t + 1 = 0

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

27 x^{2} = 27

3 (x - 6) = 5 x^{2} + 1

What are the solutions to the equation x^2+(-3/6)^2=1 ?
The solutions to the equation x^2+(-3/6)^2=1 are x=(sqrt(3))/2 ,x=-(sqrt(3))/2
Find the zeros of x^2+(-3/6)^2=1
The zeros of x^2+(-3/6)^2=1 are x=(sqrt(3))/2 ,x=-(sqrt(3))/2