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

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

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

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

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

x = 29, x = - 29

Solution steps

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

Switch sides

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

Simplify

2 1^{2} + 2 0^{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

x (2 x - 3) = 2

12 x^{2} - 12 x + 4 = 0

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

x^{2} + 10 x - 100 = 0

4 0^{2} + 5 0^{2} = c^{2}

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