# SOLUTION: I'm not sure if I answered this problem right: The square of the first consecutive integer plus the square of the third consecutive integer equals 486 more than the square

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 Click here to see ALL problems on Square-cubic-other-roots Question 39405: I'm not sure if I answered this problem right: The square of the first consecutive integer plus the square of the third consecutive integer equals 486 more than the square of the consecutive integer. find the integers. Here's my work: x^2 +x^2 +4X + 4 = 486 + x^2 +2x +1 2x^2 + 4x +4 = 487 + x^2 + 2x Subtract x^2 + 2x + 487 from each side x^2 + 2x - 483 = 0 (x+23) (x-21) x = -23 x=21Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!GOOD TO SEE YOUR ATTEMP.KEEP IT UP!SEE MY COMMENTS BELOW ----------------------------------------------------------------------- I'm not sure if I answered this problem right: The square of the first consecutive integer plus the square of the third consecutive integer equals 486 more than the square of the consecutive integer. find the integers. Here's my work: x^2 + (x+2)^2 = 486 + (x+1)^2 x^2 +x^2 +4X + 4 = 486 + x^2 +2x +1 2x^2 + 4x +4 = 487 + x^2 + 2x Subtract x^2 + 2x + 487 from each side x^2 + 2x - 483 = 0 (x+23) (x-21) x = -23 x=21..EXCELLENT ! NOTHING BETTER COULD BE DONE.KEEP IT UP SO THE INTEGERS ARE 21,22,23 ................OR.......-23,-22,-21