jmc

Applying the distributive property twice on the left-hand side gives $$x(x^2-4x+3)+5(x^2-4x+3)-x(x^2+4x-5)+c(x^2+4x-5)=0.$$ Simplifying by expanding each product and collecting like powers of $x$ gives us $$(c-3)x^2+(4c-12)x+(15-5c)=0.$$ The only value of $c$ for which this equation is always true for all $x$ is $c=\boxed{3}$.