jmc

By Vieta's formulas, we know that $$\begin{array}{rl}a+b+c& =\frac{1}{2},\\ ab+bc+ca& =\frac{4}{2}=2,\\ abc& =-\frac{10}{2}=-5.\end{array}$$We square both sides of $a+b+c=\frac{1}{2},$ which will produce the terms $a^2+b^2+c^2$: $$(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca=\frac{1}{4}.$$Substituting $ab+bc+ca=2,$ we have $$a^2+b^2+c^2+2(2)=\frac{1}{4},$$so $$a^2+b^2+c^2=\frac{1}{4}-4=\boxed{-\frac{15}{4}}.$$