What are the solutions to the equation x^2= 25/289 ?

The solutions to the equation x^2= 25/289 are x= 5/17 ,x=-5/17

Find the zeros of x^2= 25/289

The zeros of x^2= 25/289 are x= 5/17 ,x=-5/17

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x^{2} = \frac{25}{289}

x = \frac{5}{17}, x = - \frac{5}{17}

Solution steps

x^{2} = \frac{25}{289}

For

x^{2} = f (a)

the solutions are

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

x = \frac{25}{289}, x = - \frac{25}{289}

\frac{25}{289} = \frac{5}{17}

- \frac{25}{289} = - \frac{5}{17}

x = \frac{5}{17}, x = - \frac{5}{17}

2 x^{2} - 20 x = - 49

y^{2} + 75 = - 28 y

x, f (x + h) = 3 - 5 x - 2 x^{2}

10 x^{2} + 11 x - 90 = 0

x, f (x + 2) = 3 x^{2} + 4 x - 2

What are the solutions to the equation x^2= 25/289 ?
The solutions to the equation x^2= 25/289 are x= 5/17 ,x=-5/17
Find the zeros of x^2= 25/289
The zeros of x^2= 25/289 are x= 5/17 ,x=-5/17