Question

Find half of the sum of natural numbers whose squares diminished by $84$ is equal to thrice of $8$ more than the given number.

Answer 1

Let x be a natural number,
As per the condition given in question the following equation can be formed:
$\Rightarrow x^2-84=3x+24$
$\Rightarrow x^2-3x-108=0$
$\Rightarrow x^2-12x+9x-108=0$
$\Rightarrow x(x-12)+9(x-12)=0$
$\Rightarrow (x-12)(x+9)=0$
$x=12,x=-9$

$x$ can't be negative i.e.$-9$ as it is natural number

So,

$x=12$

Ans: $\frac{12}{2}=6$