The roots of ax$^{2}$+bx+c=0 are (-b+$\sqrt{b^{2} -4ac}$)/2a and (-b-$\sqrt{b^{2} -4ac}$)/2a. And one of the root is (3+2$\sqrt{2}$). The value (3+2$\sqrt{2}$) is equal to (-b+$\sqrt{b^{2} -4ac}$)/2a because sign of irrational part of both are same. ($\sqrt{2}$ is irrational)
Compare rational part and irrational part separately,
(-b+$\sqrt{b^{2} -4ac}$)/2a = (3+2$\sqrt{2}$)
=> (-b+$\sqrt{b^{2} -4ac}$) = (6a+$\sqrt{32a^{2}}$) => b = -6a & c = a (b$^{2}$-4ac = 32a$^{2}$ => 36a$^{2}$ – 4ac = 32a$^{2}$ => c = a).
Similarly we have to do for 2nd euqation :-
The roots of ax$^{2}$+mx+n=0 are (-m+$\sqrt{m^{2} -4an}$)/2a and (-m-$\sqrt{m^{2} -4an}$)/2a. And one of the root is (4+2$\sqrt{3}$). The value (4+2$\sqrt{3}$) is equal to (-m+$\sqrt{m^{2} -4an}$)/2a because sign of irrational part of both are same. ($\sqrt{3}$ is irrational)
Compare rational part and irrational part separately,
(-m+$\sqrt{m^{2} -4an}$)/2a = (4+2$\sqrt{3}$)
=> (-m+$\sqrt{m^{2} -4an}$) = (8a+$\sqrt{48a^{2}}$) => m = -8a & n = 4a (m$^{2}$-4an = 48a$^{2}$ => 64a$^{2}$ – 4an = 48a$^{2}$ => n = 4a).
Therefore the value of (b/m)+(c-2b)/n = (-6a/-8a)+(a+12a)/4a = 4.