What are the solutions to the equation 20^2+b^2=29^2 ?

The solutions to the equation 20^2+b^2=29^2 are b=21,b=-21

Find the zeros of 20^2+b^2=29^2

The zeros of 20^2+b^2=29^2 are b=21,b=-21

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2 0^{2} + b^{2} = 2 9^{2}

b = 21, b = - 21

Solution steps

2 0^{2} + b^{2} = 2 9^{2}

Move

2 0^{2}

to the right side

b^{2} = 441

For

x^{2} = f (a)

the solutions are

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

b = 441, b = - 441

441 = 21

- 441 = - 21

b = 21, b = - 21

c - 12 > - 16

x^{2} - 7 x + 11 = 1

3 x^{2} - 3 x + 1 = 0

- \infty < y < \infty

0 = 40 - 490 t^{2}

What are the solutions to the equation 20^2+b^2=29^2 ?
The solutions to the equation 20^2+b^2=29^2 are b=21,b=-21
Find the zeros of 20^2+b^2=29^2
The zeros of 20^2+b^2=29^2 are b=21,b=-21