Hi Ashish,
We are using the Pythagoras theorem as it is a right-angled triangle, and we can write the sum of the squares of the sides equal to the square of the hypotenuse. 
In the calculation of the equation, $$\left(2x\right)^2\ +\ \left(2x+\ 12\right)^2\ =\ 60^2$$
we can write this equation as,
$$\left(2\times\ x\right)^2\ +\ \left(2\left(x+\ 6\right)\right)^2\ =\ \left(2\ \times\ 30\right)^2$$
$$4\left(\ x\right)^2\ +\ 4\left(x+\ 6\right)^2\ =\ 4\left(30\right)^2$$
$$4\left(x^2\ +\ \left(x+\ 6\right)^2\ \right)=\ 4\left(30\right)^2$$
$$x^2\ +\ \left(x+\ 6\right)^2\ =\ 30^2$$
Hope this helps!