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i tried so hard but i did not get the ans.

three cosecutive natural numbers are such that the square of the middle nuber exceeds the difference of the squares of the other two by 60.
assume the middle number to be x and form an quadratic equation satisfying the above statement . Hence find the three numbers.

Since the numbers are consecutive, they can be written as

$x-1, \;\ x, \;\ x+1$

The square of the middle number is $x^{2}$

The difference of the squares of the other two:

$(x+1)^{2}-(x-1)^{2}$

So, since the square of the middle is 60 larger than the difference of the squares of the other two, we have:

$x^{2}=(x+1)^{2}-(x-1)^{2}+60$

Solve for x.

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