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

The solutions to the equation 12^2+b^2=20^2 are b=16,b=-16

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

The zeros of 12^2+b^2=20^2 are b=16,b=-16

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

b = 16, b = - 16

Solution steps

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

Move

1 2^{2}

to the right side

b^{2} = 256

For

x^{2} = f (a)

the solutions are

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

b = 256, b = - 256

256 = 16

- 256 = - 16

b = 16, b = - 16

x + 5 > 7

(\frac{2}{3}) (\frac{9}{5}) (\frac{5}{4})

∣ 3 x - 4 ∣ < 8

a^{2} + 1 5^{2} = 1 7^{2}

\frac{5 ^{12} \cdot c ^{10} \cdot d ^{2}}{5 ^{9} \cdot c ^{9}}

What are the solutions to the equation 12^2+b^2=20^2 ?
The solutions to the equation 12^2+b^2=20^2 are b=16,b=-16
Find the zeros of 12^2+b^2=20^2
The zeros of 12^2+b^2=20^2 are b=16,b=-16