What are the solutions to the equation 14^2+x^2=20^2 ?

The solutions to the equation 14^2+x^2=20^2 are x=2sqrt(51),x=-2sqrt(51)

Find the zeros of 14^2+x^2=20^2

The zeros of 14^2+x^2=20^2 are x=2sqrt(51),x=-2sqrt(51)

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1 4^{2} + x^{2} = 2 0^{2}

x = 251, x = - 251

Decimal Notation

x = 14.28285 \dots, x = - 14.28285 \dots

Solution steps

1 4^{2} + x^{2} = 2 0^{2}

Move

1 4^{2}

to the right side

x^{2} = 204

For

x^{2} = f (a)

the solutions are

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

x = 204, x = - 204

204 = 251

- 204 = - 251

x = 251, x = - 251

\frac{3}{x + 2} - \frac{5}{x + 9} \geq - 1

- 2 \cdot - 6

9 w^{2} - 30 w = - 25

9 (x - 7) = - 3 (x + 53)

3 x - 4 > 8

What are the solutions to the equation 14^2+x^2=20^2 ?
The solutions to the equation 14^2+x^2=20^2 are x=2sqrt(51),x=-2sqrt(51)
Find the zeros of 14^2+x^2=20^2
The zeros of 14^2+x^2=20^2 are x=2sqrt(51),x=-2sqrt(51)