解；
方程有实根,判别式≥0
(m-2)²-4(m²+3m+5)≥0
3m²+16m+16≤0
(m+4)(3m+4)≤0
-4≤m≤-4/3
设两根分别为x1,x2.由韦达定理得
x1+x2=2-m
x1x2=m²+3m+5
x1²+x2²=(x1+x2)²-2x1x2
=(2-m)²-2(m²+3m+5)
=-m²-10m-6
=-(m+5)² +19
又-4≤m≤-4/3
当m=-4时,x1²+x2²有最大值(x1²+x2²)max=18.