What are the solutions to the equation 21^2+20^2=c^2 ?

The solutions to the equation 21^2+20^2=c^2 are c=29,c=-29

Find the zeros of 21^2+20^2=c^2

The zeros of 21^2+20^2=c^2 are c=29,c=-29

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 1^{2} + 2 0^{2} = c^{2}

c = 29, c = - 29

Solution steps

2 1^{2} + 2 0^{2} = c^{2}

Switch sides

c^{2} = 2 1^{2} + 2 0^{2}

Simplify

2 1^{2} + 2 0^{2} : 841

For

x^{2} = f (a)

the solutions are

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

c = 841, c = - 841

841 = 29

- 841 = - 29

c = 29, c = - 29

(- 4 b^{3}) (2 b^{5})

x^{2} - 4 x + 2 = 7

\frac{1}{2} x - \frac{3}{5} < 2 - \frac{7}{10} x

0.3 \cdot 4

6 + 17 y = - 15 + 14 y

What are the solutions to the equation 21^2+20^2=c^2 ?
The solutions to the equation 21^2+20^2=c^2 are c=29,c=-29
Find the zeros of 21^2+20^2=c^2
The zeros of 21^2+20^2=c^2 are c=29,c=-29