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

The solutions to the equation 2^2+b^2=11^2 are b=3sqrt(13),b=-3sqrt(13)

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

The zeros of 2^2+b^2=11^2 are b=3sqrt(13),b=-3sqrt(13)

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

b = 313, b = - 313

Decimal Notation

b = 10.81665 \dots, b = - 10.81665 \dots

Solution steps

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

Move

2^{2}

to the right side

b^{2} = 117

For

x^{2} = f (a)

the solutions are

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

b = 117, b = - 117

117 = 313

- 117 = - 313

b = 313, b = - 313

y, x y - 1 = x + y^{2}

(k - 2) x^{2} - 5 x + 2 k = 0.6

3 x^{2} + 2 x - 33 = 0

- 3 = 6 x + 5 x^{2}

x, x^{2} + 2 x = 8

What are the solutions to the equation 2^2+b^2=11^2 ?
The solutions to the equation 2^2+b^2=11^2 are b=3sqrt(13),b=-3sqrt(13)
Find the zeros of 2^2+b^2=11^2
The zeros of 2^2+b^2=11^2 are b=3sqrt(13),b=-3sqrt(13)