Find two positive numbers whose squares have the difference $48$ and the sum of the numbers is $12$.

Solve for $x:$

$a\left(a^2+b^2\right)x^2+b^{2}x-a=0$

The numerator of fraction is $2$ less than its denominator. If $3$ is added to both numerator as well as denominator, the sum of new and original fraction is $\frac{29}{18},$ find the fraction.