What are the solutions to the equation 3^2+b^2=4^2 ?

The solutions to the equation 3^2+b^2=4^2 are b=sqrt(7),b=-sqrt(7)

Find the zeros of 3^2+b^2=4^2

The zeros of 3^2+b^2=4^2 are b=sqrt(7),b=-sqrt(7)

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3^{2} + b^{2} = 4^{2}

b = 7, b = - 7

Decimal Notation

b = 2.64575 \dots, b = - 2.64575 \dots

Solution steps

3^{2} + b^{2} = 4^{2}

Move

3^{2}

to the right side

b^{2} = 7

For

x^{2} = f (a)

the solutions are

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

b = 7, b = - 7

x^{2} - 6 x - 8 = 0

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

\frac{7 p + 4}{8} = 3

x (x + 9) = 0

2 x^{2} + 4 x - 3 = 0

What are the solutions to the equation 3^2+b^2=4^2 ?
The solutions to the equation 3^2+b^2=4^2 are b=sqrt(7),b=-sqrt(7)
Find the zeros of 3^2+b^2=4^2
The zeros of 3^2+b^2=4^2 are b=sqrt(7),b=-sqrt(7)