SOLUTION: X^2+(x+6)^2=(x+12)^2 I got the answer with trial an error. But i don't understand how to show the work. (Finding the lengths of all legs of a right triangle)

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Question 982865: X^2+(x+6)^2=(x+12)^2
I got the answer with trial an error. But i don't understand how to show the work.
(Finding the lengths of all legs of a right triangle)
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
X^2+(x+6)^2=(x+12)^2
I got the answer with trial an error. But i don't understand how to show the work.
(Finding the lengths of all legs of a right triangle)
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X^2+(x+6)^2=(x+12)^2
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x^2 + x^2 + 12x + 36 = x^2 + 24x + 144
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x^2 - 12x - 108 = 0
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(x-18)(x+6) = 0
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Positive solution::
x = 18
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Cheers,
Stan H.
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Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
X^2 + (x+6)^2 = (x+12)^2
FOIL (x+6)(x+6) and (x+12)(x+12)
x^2 + x^2 + 12x + 36 = x^2 + 24x + 144
Collect like terms on the left
x^2 + x^2 - x^2 + 12x - 24x + 36 - 144 = 0
x^2 - 12x - 108 = 0
You can use the quadratic formula a=1; b=-12, c=-108; but this will factor to
(x - 18)(x + 6) = 0
It's the positive solution we want here
x = 18 is one side
then
18 + 6 = 24 is the other side
and
18 + 12 = 30 is the hypotenuse