What are the solutions to the equation b^2+11^2=21^2 ?

The solutions to the equation b^2+11^2=21^2 are b=8sqrt(5),b=-8sqrt(5)

Find the zeros of b^2+11^2=21^2

The zeros of b^2+11^2=21^2 are b=8sqrt(5),b=-8sqrt(5)

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

b = 85, b = - 85

Decimal Notation

b = 17.88854 \dots, b = - 17.88854 \dots

Solution steps

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

Move

1 1^{2}

to the right side

b^{2} = 320

For

x^{2} = f (a)

the solutions are

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

b = 320, b = - 320

320 = 85

- 320 = - 85

b = 85, b = - 85

- 10 + 3 x = 5

y, 3 x + 5 y = 50

- \frac{1}{4} x \leq 2

(\frac{9}{7})^{- 1}

\frac{9 ^{- 2}}{3 ^{- 3}}

What are the solutions to the equation b^2+11^2=21^2 ?
The solutions to the equation b^2+11^2=21^2 are b=8sqrt(5),b=-8sqrt(5)
Find the zeros of b^2+11^2=21^2
The zeros of b^2+11^2=21^2 are b=8sqrt(5),b=-8sqrt(5)