jmc

We expand both sides to find

$$\begin{array}{rl}(w+13)(w+13)& =(3w+7)(2w+4)\\ w^2+26w+169& =3w(2w+4)+7(2w+4)\\ w^2+26w+169& =6w^2+12w+14w+28\\ w^2+26w+169& =6w^2+26w+28\\ w^2+169& =6w^2+28\\ 141& =5w^2\\ \frac{141}{5}& =w^2.\end{array}$$

So, expressed as a decimal, our answer is $\frac{141}{5}=\boxed{28.2}$.