What are the solutions to the equation 2^2+b^2=3^2 ?

The solutions to the equation 2^2+b^2=3^2 are b=sqrt(5),b=-sqrt(5)

Find the zeros of 2^2+b^2=3^2

The zeros of 2^2+b^2=3^2 are b=sqrt(5),b=-sqrt(5)

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2^{2} + b^{2} = 3^{2}

b = 5, b = - 5

Decimal Notation

b = 2.23606 \dots, b = - 2.23606 \dots

Solution steps

2^{2} + b^{2} = 3^{2}

Move

2^{2}

to the right side

b^{2} = 5

For

x^{2} = f (a)

the solutions are

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

b = 5, b = - 5

s^{2} + 7 s = - 17

- 2 = 3 x^{2} - x - 2

0 \leq 2 x - 6 \leq 4

y^{2} - 25 x^{2}

2 x - x 5 = 9

What are the solutions to the equation 2^2+b^2=3^2 ?
The solutions to the equation 2^2+b^2=3^2 are b=sqrt(5),b=-sqrt(5)
Find the zeros of 2^2+b^2=3^2
The zeros of 2^2+b^2=3^2 are b=sqrt(5),b=-sqrt(5)