What are the solutions to the equation a^2+20^2=27^2 ?

The solutions to the equation a^2+20^2=27^2 are a=sqrt(329),a=-sqrt(329)

Find the zeros of a^2+20^2=27^2

The zeros of a^2+20^2=27^2 are a=sqrt(329),a=-sqrt(329)

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a^{2} + 2 0^{2} = 2 7^{2}

a = 329, a = - 329

Decimal Notation

a = 18.13835 \dots, a = - 18.13835 \dots

Solution steps

a^{2} + 2 0^{2} = 2 7^{2}

Move

2 0^{2}

to the right side

a^{2} = 329

For

x^{2} = f (a)

the solutions are

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

a = 329, a = - 329

∣ 3 x + 2 ∣ \geq 5

- \frac{6}{6}

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

5 n - 8 > 6

4 (x - 3) = 5 x + 2

What are the solutions to the equation a^2+20^2=27^2 ?
The solutions to the equation a^2+20^2=27^2 are a=sqrt(329),a=-sqrt(329)
Find the zeros of a^2+20^2=27^2
The zeros of a^2+20^2=27^2 are a=sqrt(329),a=-sqrt(329)