If $$(3+2\sqrt{2})$$ is a root of the equation $$ax^{2}+bx+c=0$$ and $$(4+2\sqrt{3})$$ is a root of the equation $$ay^{2}+my+n=0$$ where a, b, c, m and n are integers, then the value of $$(\frac{b}{m}+\frac{c-2b}{n})$$ is

a, b, c, m and n are integers so if one root is $$3+2\sqrt{2}$$ then the other root is $$3-2\sqrt{2}$$

Sum of roots = 6 = -b/a  or b= -6a

Product of roots = 1 = c/a  or c=a

a, b, c, m and n are integers so if one root is $$4+2\sqrt{3}$$ then the other root is $$4-2\sqrt{3}$$

Sum of roots = 8 = -m/a or m = -8a

product of roots = 4 = n/a or n = 4a

$$(\frac{b}{m}+\frac{c-2b}{n})$$

= $$\frac{6a}{8a}+\frac{\left(a+12a\right)}{4a}=\frac{3}{4}+\frac{13}{4}=\frac{16}{4}=4$$