What are the solutions to the equation 4^2+x^2=sqrt(20)^2 ?

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

Find the zeros of 4^2+x^2=sqrt(20)^2

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

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4^{2} + x^{2} = 20^{2}

x = 2, x = - 2

Solution steps

4^{2} + x^{2} = (20)^{2}

Move

4^{2}

to the right side

x^{2} = 4

For

x^{2} = f (a)

the solutions are

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

x = 4, x = - 4

4 = 2

- 4 = - 2

x = 2, x = - 2

3 x - 5 > - 8

(2 x + 1)^{2} - 3 (2 x + 1) - 10 = 0

\frac{4}{9} = \frac{r - 3}{6}

x - \frac{1}{3} \leq \frac{7}{12}

\frac{7}{3} - \frac{32}{3}

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