jmc

Let $x=a-5.$ Then $a=x+5,$ so $$(x+5{)}^3-15(x+5{)}^2+20(x+5)-50=0,$$which simplifies to $x^3-55x-200=0.$

Let $y=b-\frac{5}{2}.$ Then $b=y+\frac{5}{2},$ so $$8{\left(y+\frac{5}{2}\right)}^3-60{\left(y+\frac{5}{2}\right)}^2-290\left(y+\frac{5}{2}\right)+2575=0,$$which simplifies to $y^3-55y+200=0.$ (Note that through these substitutions, we made the quadratic term vanish in each of these cubic equations.)

Consider the function $f(t)=t^3-55t.$ Observe that the polynomial $f(t)$ has three roots 0, $\sqrt{55},$ and $-\sqrt{55}.$ Its graph is shown below.



Let $0\le t\le \sqrt{55}.$ Then $$[f(t){]}^2=(t^3-55t{)}^2=t^2(t^2-55{)}^2=t^2(55-t^2{)}^2=t^2(55-t^2)(55-t^2).$$By AM-GM, $$2t^2(55-t^2)(55-t^2)\le {\left(\frac{(2t^2)+(55-t^2)+(55-t^2)}{3}\right)}^3={\left(\frac{110}{3}\right)}^3<40^3,$$so $$[f(t){]}^2<32000<32400,$$which means $|f(t)|<180.$

Since $f(t)$ is an odd function, $|f(t)|<180$ for $-\sqrt{55}\le t\le 0$ as well. This means that the equation $f(t)=200$ has exactly one real root. Similarly, $f(t)=-200$ has exactly one real root. Furthermore, since $f(t)$ is an odd function, these roots add up to 0.

Then $$a-5+b-\frac{5}{2}=0,$$so $a+b=5+\frac{5}{2}=\boxed{\frac{15}{2}}.$