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

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

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

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

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

x = 5, x = - 5

Decimal Notation

x = 2.23606 \dots, x = - 2.23606 \dots

Solution steps

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

Switch sides

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

Move

2^{2}

to the right side

x^{2} = 5

For

x^{2} = f (a)

the solutions are

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

x = 5, x = - 5

(x - \frac{\frac{7}{6}}{2})^{2} - (\frac{\frac{- 7}{6}}{2})^{2} - \frac{1}{4} = 0

(x + 3)^{2} - 2 = 2 x (x + 2) + x + 8

x^{2} + 71 = 18 x - 9

m^{2} - 3 m + 5 = 0

(y + 7)^{2} = 2 y^{2} + 17 + 39

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