What are the solutions to the equation 20^2+21^2=x^2 ?

The solutions to the equation 20^2+21^2=x^2 are x=29,x=-29

Find the zeros of 20^2+21^2=x^2

The zeros of 20^2+21^2=x^2 are x=29,x=-29

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2 0^{2} + 2 1^{2} = x^{2}

x = 29, x = - 29

Solution steps

2 0^{2} + 2 1^{2} = x^{2}

Switch sides

x^{2} = 2 0^{2} + 2 1^{2}

Simplify

2 0^{2} + 2 1^{2} : 841

For

x^{2} = f (a)

the solutions are

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

x = 841, x = - 841

841 = 29

- 841 = - 29

x = 29, x = - 29

(\frac{10 - x}{4})^{2}

x^{3} - y^{3} - x^{2} y + x y^{2}

\frac{3}{2} x - 1 > 8 x - \frac{7}{6}

1 - \frac{1}{36}

- 2 (z - 1) \leq 18

What are the solutions to the equation 20^2+21^2=x^2 ?
The solutions to the equation 20^2+21^2=x^2 are x=29,x=-29
Find the zeros of 20^2+21^2=x^2
The zeros of 20^2+21^2=x^2 are x=29,x=-29