The product of $2$ consecutive no.is greater than the square of smaller no. by $30$

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Solution

Let number be $a, a + 1$
$a (a + 1) = a^{2} + 30 a^{2} + a = a^{2} + 30 a = 30$

So, numbers are $30, 31$ .

Hence, this is the answer.

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The product of 2 consecutive no.is greater than the square of smaller no. by 30

Let number be a,a+1a(a+1)=a2+30a2+a=a2+30a=30So, numbers are 30,31.Hence, this is the answer.

The product of $2$ consecutive no.is greater than the square of smaller no. by $30$

Let number be $a, a + 1$
$a (a + 1) = a^{2} + 30 a^{2} + a = a^{2} + 30 a = 30$

So, numbers are $30, 31$ .

Hence, this is the answer.