SOLUTION: Find two consecutive integers such that the square of the second added to the square of twice the first is 424.

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Question 522022: Find two consecutive integers such that the square of the second added to the square of twice the first is 424.
Answer by Maths68(1474) About Me  (Show Source):
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Let
Ist Integer = x
2nd Integer = x+1
%28x%2B1%29%5E2%2B%282x%29%5E2=424
x%5E2%2B2x%2B1%2B%282x%29%5E2=424
x%5E2%2B2x%2B1%2B4x%5E2=424
5x%5E2%2B2x%2B1=424
5x%5E2%2B2x%2B1-424=0
5x%5E2%2B2x-423=0
Use
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

a+=+5
b+=+2
c+=+-423

x+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A5%2A-423+%29%29%2F%282%2A5%29+

x+=+%28-2+%2B-+sqrt%28+4%2B8460+%29%29%2F%2810%29+

x+=+%28-2+%2B-+sqrt%288464%29%29%2F%2810%29+

x+=+%28-2+%2B-+92%29%2F%2810%29+

x+=+%28-2+-+92%29%2F%2810%29+ or x+=+%28-2+%2B+92%29%2F%2810%29+

x+=+%28-94%29%2F%2810%29+ or x+=+%2890%29%2F%2810%29+

x=-9.4 or x=9%29+
x=-9.4 inadmissible
x=9%29+

Ist Integer = x = 9
2nd Integer = x+1 = 9+1 = 10