Least Value

Simply we can say that,

the roots are $\alpha$ and $\beta$ .

So By Veita's Formula

$\alpha + \beta = 2-m$

$\alpha \beta = -(m+3)$

Solving for $\alpha ^2+ \beta ^2$ we get

$\alpha ^2+ \beta ^2 = m^2-2m+1$

Now this sum is minimum when

$\dfrac{d(\alpha ^2+ \beta ^2)}{dm} = 0$

So We get $2m-2 = 0$

So $\boxed{m=1}$

Let the roots be $A, B$ and by Vieta's Formulae:

$A+B = -(m-2) (i), AB = -(m+3) (ii)$

Squaring (i) out produces $A^2 + 2AB + B^2 = (m-2)^2 \Rightarrow A^2 + B^2 = (m-2)^2 - 2(-m-3) = m^2 -2m + 10 = (m-1)^2 + 9 (iii)$

Hence, (iii) is just a concave-up parabola that attains a global minimum value of $9$ when $\boxed{m=1}.$