What are the solutions to the equation 2^2+sqrt(21)^2=x^2 ?

The solutions to the equation 2^2+sqrt(21)^2=x^2 are x=5,x=-5

Find the zeros of 2^2+sqrt(21)^2=x^2

The zeros of 2^2+sqrt(21)^2=x^2 are x=5,x=-5

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2^{2} + 21^{2} = x^{2}

x = 5, x = - 5

Solution steps

2^{2} + (21)^{2} = x^{2}

Switch sides

x^{2} = 2^{2} + (21)^{2}

Simplify

2^{2} + (21)^{2} : 25

For

x^{2} = f (a)

the solutions are

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

x = 25, x = - 25

25 = 5

- 25 = - 5

x = 5, x = - 5

9 x^{2} - 6 x - 7 = 0

(x - 1)^{2} = 121

a, a^{2} = b^{2} + c^{2}

x^{2} + 8 x + 18 = 3

3 x^{2} - 1 = 4 x

What are the solutions to the equation 2^2+sqrt(21)^2=x^2 ?
The solutions to the equation 2^2+sqrt(21)^2=x^2 are x=5,x=-5
Find the zeros of 2^2+sqrt(21)^2=x^2
The zeros of 2^2+sqrt(21)^2=x^2 are x=5,x=-5