14, 21
Explanation:
Let the two consecutive multiples be 7n and 7(n+1).
Now, it is given that:
(7n)² + [7(n+1)]² = 637
Simplifying,
49n² + (7n+7)² = 637
49n² + 49n² + 2(7n)(7) + 49 = 637
(as (x+y)² = x² + y² + 2xy)
98n² + 98n + 49 = 637
98n² + 98n - 588 = 0
Dividing by 98 to further simplify,
n² + n - 6 = 0
Factorising, we get
n² + 3n - 2n - 6 = 0
n(n+3) - 2(n+3) = 0
(n+3)(n-2) = 0
Therefore we get n = -3 or n = +2.
Hence, we get the consecutive multiples as 14 and 21.
Feel free to ask if you didn't understand any part.
Hope this helps! :D