Standing Point Of Exponential Function Combined With Polynomial

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How to Find Minimum Of the below function using Calculus Method?

$$f(x)=(x-2)e^{x-2}-(x+3)^2$$

After Differentiating Once, the equation turns out to be:

$$f '(x)= -2e^{x-2} + 2x.e^{x-2} - 2x -6.$$

From this step to find the minimum, we need to find the zeros of $f '(x)$. But how to find the zeros of the equation $f '(x)$ with exponentials and polynomial combined together? Should I Use Logarithm to find zeros of such equations?

Though we can find the minimum using Numerical methods, I am interested in Knowing the Calculus way of solving This.

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Considering that you search for the minimum of function $$f(x)=(x-2)e^{x-2}-(x+3)^2$$ $$f'(x)=e^{x-2} (x-1)-2 (x+3)=0 \implies e^{x-2}=2\,\frac{x+3 } {x-1 }$$ Make $x-2=-t$ to get $$e^{-t}=2\,\frac{ t-5}{t-1}$$ Have a look at this paper and in particular at equation $(4)$. The solution is "just" given in terms of the generalized Lambert function.

From a practical point of view, you will need some numerical method.