What are the solutions to the equation 10^2+b^2=26^2 ?

The solutions to the equation 10^2+b^2=26^2 are b=24,b=-24

Find the zeros of 10^2+b^2=26^2

The zeros of 10^2+b^2=26^2 are b=24,b=-24

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1 0^{2} + b^{2} = 2 6^{2}

b = 24, b = - 24

Solution steps

1 0^{2} + b^{2} = 2 6^{2}

Move

1 0^{2}

to the right side

b^{2} = 576

For

x^{2} = f (a)

the solutions are

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

b = 576, b = - 576

576 = 24

- 576 = - 24

b = 24, b = - 24

\frac{x ^{2} - y ^{2}}{y - x}

20.5 x^{2} + 567 x + 3892.5 = 0

\frac{6 c ^{3}}{3 c ^{- 5}}

x^{3} - 2 x^{2} - 25 x + 50

What are the solutions to the equation 10^2+b^2=26^2 ?
The solutions to the equation 10^2+b^2=26^2 are b=24,b=-24
Find the zeros of 10^2+b^2=26^2
The zeros of 10^2+b^2=26^2 are b=24,b=-24