Let the required consecutive multiples of 7 be 7x and 7(x + 1)

According to given condition,

(7x)² + [7(x + 1)]² = 1225

49 x² + 49(x² + 2x + 1) = 1225 using (a + b)² = a² + 2ab + b²

49 x² + 49x² + 98x + 49 = 1225

98x² + 98x–1176 = 0

x² + x – 12 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 1 c = – 12

= 1. – 12 = – 12

And either of their sum or difference = b

= 1

Thus the two terms are 4 and – 3

Difference = 4 – 3 = 1

Product = 4. – 3 = – 12

x² + 4x – 3x – 12 = 0

x(x + 4) – 3(x + 4) = 0

(x – 3) (x + 4) = 0

(x – 3) = 0 or (x + 4) = 0

x = 3 or x = – 4

when x = 3,

7x = 7.3 = 21

7(x + 1) = 7(3 + 1) = 7.4 = 28

Hence required multiples are 21, 28.