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

The solutions to the equation x^2=4^2+2^2 are x=2sqrt(5),x=-2sqrt(5)

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

The zeros of x^2=4^2+2^2 are x=2sqrt(5),x=-2sqrt(5)

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

x = 25, x = - 25

Decimal Notation

x = 4.47213 \dots, x = - 4.47213 \dots

Solution steps

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

Simplify

4^{2} + 2^{2} : 20

For

x^{2} = f (a)

the solutions are

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

x = 20, x = - 20

20 = 25

- 20 = - 25

x = 25, x = - 25

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

8 x^{2} - 5 x + 3 = 0

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

- 5 x^{2} + 12 x + 7 = 0

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

What are the solutions to the equation x^2=4^2+2^2 ?
The solutions to the equation x^2=4^2+2^2 are x=2sqrt(5),x=-2sqrt(5)
Find the zeros of x^2=4^2+2^2
The zeros of x^2=4^2+2^2 are x=2sqrt(5),x=-2sqrt(5)