jmc

Setting the $y$-values to be equal, we get $$x^4-5x^2-x+4=x^2-3x,$$so $x^4-6x^2+2x+4=0.$ Let the four roots of this polynomial be $a,$ $b,$ $c,$ and $d.$ Then by Vieta's formulas, $$\begin{array}{rl}a+b+c+d& =0,\\ ab+ac+ad+bc+bd+cd& =-6.\end{array}$$We want the sum of the $y$-values, which is $$(a^2-3a)+(b^2-3b)+(c^2-3c)+(d^2-3d)=(a^2+b^2+c^2+d^2)-3(a+b+c+d)=a^2+b^2+c^2+d^2.$$Squaring the equation $a+b+c+d=0,$ we get $$a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)=0.$$Then $$a^2+b^2+c^2+d^2=-2(ab+ac+ad+bc+bd+cd)=\boxed{12}.$$