Question 468090
The square of the first consecutive plus the square of the third consecutive integer equals 486 more than the square of the second consecutive integers.

I TRY:

x^2(x+2)^2=(x+1)^2+486 (It was a mess when factor everything out)
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It says plus, you multiplied
x^2 + (x+2)^2=(x+1)^2+486
{{{x^2 + x^2 + 4x + 4 = x^2 + 2x + 487}}}
{{{x^2 + 2x - 483 = 0}}}
(x - 21)*(x + 23 = 0
--> 21 & 23 
or -23 & -21