$\begin{array}{l}\text{Given that the equation }{x}^2-\left(k-2\right)x+\left(k^2+3k+5\right)=0\text{\hspace{0.17em}has real roots}\\ \text{it means its discriminant must be positive }\\ {\left(k+2\right)}^2-4\left(k^2+3k+5\right)\ge 0\\ -3k^2-16k-16\ge 0\\ 3k^2+16k+16\le 0\\ \left(3k+4\right)\left(k+4\right)\le 0\\ k\in \left[-4,-\frac{4}{3}\right]\end{array}$

$\begin{array}{rcl}Consider {\alpha }^2+{\beta }^2& =& {\left(\alpha +\beta \right)}^2-2\alpha \beta \\ & =& {\left(k-2\right)}^2-2\left(k^2+3k+5\right)\\ & =& -k^2-10k-6\\ & & \end{array}$

Differentiate both sides and equate to zero

$$\begin{gathered} 2k=-10 \\ k=-5 \end{gathered}$$

but this value is not in the range of k values 

hence the minimum or maximum values of $\begin{array}{rcl} & & \end{array}\begin{array}{rcl} & & \alpha \end{array}\begin{array}{rcl} & & \end{array}$ are at $k=-4,-\frac{4}{3}$

Therefore, the extremum values are $18,\frac{50}{9}$