What are the solutions to the equation 12^2+20^2=c^2 ?

The solutions to the equation 12^2+20^2=c^2 are c=4sqrt(34),c=-4sqrt(34)

Find the zeros of 12^2+20^2=c^2

The zeros of 12^2+20^2=c^2 are c=4sqrt(34),c=-4sqrt(34)

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

c = 434, c = - 434

Decimal Notation

c = 23.32380 \dots, c = - 23.32380 \dots

Solution steps

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

Switch sides

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

Simplify

1 2^{2} + 2 0^{2} : 544

For

x^{2} = f (a)

the solutions are

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

c = 544, c = - 544

544 = 434

- 544 = - 434

c = 434, c = - 434

(r + \frac{7}{8})^{2} = \frac{1}{64}

3 x^{2} + 48 x + 183 = 0

y^{2} - 5 y + 1 = 0

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

8 y^{2} + y + 2 = 0

What are the solutions to the equation 12^2+20^2=c^2 ?
The solutions to the equation 12^2+20^2=c^2 are c=4sqrt(34),c=-4sqrt(34)
Find the zeros of 12^2+20^2=c^2
The zeros of 12^2+20^2=c^2 are c=4sqrt(34),c=-4sqrt(34)