What is the answer to 32^2+x^2=(80-x)^2+40^2 ?

The answer to 32^2+x^2=(80-x)^2+40^2 is x= 218/5

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32^2+x^2=(80-x)^2+40^2

3 2^{2} + x^{2} = (80 - x)^{2} + 4 0^{2}

x = \frac{218}{5}

Decimal Notation

x = 43.6

Solution steps

3 2^{2} + x^{2} = (80 - x)^{2} + 4 0^{2}

3 2^{2} = 1024

1024 + x^{2} = (80 - x)^{2} + 4 0^{2}

4 0^{2} = 1600

1024 + x^{2} = (80 - x)^{2} + 1600

Expand

(80 - x)^{2} + 1600 : x^{2} - 160 x + 8000

1024 + x^{2} = x^{2} - 160 x + 8000

Move

1024

to the right side

x^{2} = x^{2} - 160 x + 6976

Subtract

x^{2} - 160 x

from both sides

x^{2} - (x^{2} - 160 x) = x^{2} - 160 x + 6976 - (x^{2} - 160 x)

Simplify

160 x = 6976

Divide both sides by

160

x = \frac{218}{5}

5 x + 15 = 3 x + 29

- \frac{2}{5} = \frac{x}{20}

x, 3 x - 2 y - 5 = 6 x + 4 y + 10

0 - 8 = - 4 x

a, y = a x - b

What is the answer to 32^2+x^2=(80-x)^2+40^2 ?
The answer to 32^2+x^2=(80-x)^2+40^2 is x= 218/5
Solve for x: 32^2+x^2=(80-x)^2+40^2
The solution is x= 218/5