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

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

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

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

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

x = 434, x = - 434

Decimal Notation

x = 23.32380 \dots, x = - 23.32380 \dots

Solution steps

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

Switch sides

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

Simplify

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

For

x^{2} = f (a)

the solutions are

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

x = 544, x = - 544

544 = 434

- 544 = - 434

x = 434, x = - 434

x + 3 = 9

\frac{3}{4} x = - 27

8 \cdot \frac{x}{y}

2 x^{2} - 12 x - 26 = 0

2 a^{2} + 16 a - 40

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