What are the solutions to the equation 7^2+12^2=x^2 ?

The solutions to the equation 7^2+12^2=x^2 are x=sqrt(193),x=-sqrt(193)

Find the zeros of 7^2+12^2=x^2

The zeros of 7^2+12^2=x^2 are x=sqrt(193),x=-sqrt(193)

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7^{2} + 1 2^{2} = x^{2}

x = 193, x = - 193

Decimal Notation

x = 13.89244 \dots, x = - 13.89244 \dots

Solution steps

7^{2} + 1 2^{2} = x^{2}

Switch sides

x^{2} = 7^{2} + 1 2^{2}

Simplify

7^{2} + 1 2^{2} : 193

For

x^{2} = f (a)

the solutions are

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

x = 193, x = - 193

∣ x + 4 ∣ < 1

4 x - 1 < 7 and 2 x + 8 \geq 4

\frac{4}{3} π 4^{3}

3 x < - 6

lo g_{5} (7 a^{5})

What are the solutions to the equation 7^2+12^2=x^2 ?
The solutions to the equation 7^2+12^2=x^2 are x=sqrt(193),x=-sqrt(193)
Find the zeros of 7^2+12^2=x^2
The zeros of 7^2+12^2=x^2 are x=sqrt(193),x=-sqrt(193)