What are the solutions to the equation 21^2+29^2=c^2 ?

The solutions to the equation 21^2+29^2=c^2 are c=sqrt(1282),c=-sqrt(1282)

Find the zeros of 21^2+29^2=c^2

The zeros of 21^2+29^2=c^2 are c=sqrt(1282),c=-sqrt(1282)

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2 1^{2} + 2 9^{2} = c^{2}

c = 1282, c = - 1282

Decimal Notation

c = 35.80502 \dots, c = - 35.80502 \dots

Solution steps

2 1^{2} + 2 9^{2} = c^{2}

Switch sides

c^{2} = 2 1^{2} + 2 9^{2}

Simplify

2 1^{2} + 2 9^{2} : 1282

For

x^{2} = f (a)

the solutions are

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

c = 1282, c = - 1282

3 x^{2} + 8 x + 40 = 0

x = y^{2} + 4 y, y

- x^{2} + 5 x - 2

- x^{2} + 5 x + 3

x, x^{2} y + y^{2} = k

What are the solutions to the equation 21^2+29^2=c^2 ?
The solutions to the equation 21^2+29^2=c^2 are c=sqrt(1282),c=-sqrt(1282)
Find the zeros of 21^2+29^2=c^2
The zeros of 21^2+29^2=c^2 are c=sqrt(1282),c=-sqrt(1282)