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

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

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

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

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

b = 5, b = - 5

Decimal Notation

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

Solution steps

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

Switch sides

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

- 0.6 (x - 2.5)^{2} + 8.75 = 0

p^{2} - p - 1 = 0

- 55 = - 4 x^{2} + 6 x - 1

6 x^{2} - 2 x = 140

2 c \cdot 2 c - 2 c = 12

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