What are the solutions to the equation 4^2+12^2=x^2 ?

The solutions to the equation 4^2+12^2=x^2 are x=4sqrt(10),x=-4sqrt(10)

Find the zeros of 4^2+12^2=x^2

The zeros of 4^2+12^2=x^2 are x=4sqrt(10),x=-4sqrt(10)

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4^{2} + 1 2^{2} = x^{2}

x = 410, x = - 410

Decimal Notation

x = 12.64911 \dots, x = - 12.64911 \dots

Solution steps

4^{2} + 1 2^{2} = x^{2}

Switch sides

x^{2} = 4^{2} + 1 2^{2}

Simplify

4^{2} + 1 2^{2} : 160

For

x^{2} = f (a)

the solutions are

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

x = 160, x = - 160

160 = 410

- 160 = - 410

x = 410, x = - 410

64 x^{2} - 80 x

4 = 3 x^{2} - 26 x + 44

x^{2} + 20 x + 28 = 0

4 y^{2} = 8 y + 5

- 2 x^{2} + 2 x + 9 = - 5 x

What are the solutions to the equation 4^2+12^2=x^2 ?
The solutions to the equation 4^2+12^2=x^2 are x=4sqrt(10),x=-4sqrt(10)
Find the zeros of 4^2+12^2=x^2
The zeros of 4^2+12^2=x^2 are x=4sqrt(10),x=-4sqrt(10)