What are the solutions to the equation a^2=289 ?

The solutions to the equation a^2=289 are a=17,a=-17

Find the zeros of a^2=289

The zeros of a^2=289 are a=17,a=-17

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a^{2} = 289

a = 17, a = - 17

Solution steps

a^{2} = 289

For

x^{2} = f (a)

the solutions are

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

a = 289, a = - 289

289 = 17

- 289 = - 17

a = 17, a = - 17

∣ 4 x - 6 ∣ + 2 \geq 7

27 x^{3} + 8 y^{3}

n^{2} + 2 n + n p + 2 p

\frac{f ( 3 + h ) - f ( 3 )}{h}

3 x^{2} + 12 x - 6 x^{3}

What are the solutions to the equation a^2=289 ?
The solutions to the equation a^2=289 are a=17,a=-17
Find the zeros of a^2=289
The zeros of a^2=289 are a=17,a=-17