Since the given equation is a quadratic equation, we are expecting to find two solutions and since floor function is involved, the solutions must be in a range of some numbers.
Bearing this in mind, I noticed that if I let $$\left\lfloor x+\frac{7}{3} \right\rfloor=4$$ and $$\left\lfloor x-\frac{9}{4} \right\rfloor=0$$ then I would get $$\left\lfloor x+\frac{7}{3}\right\rfloor^2+\left\lfloor x-\frac{9}{4}\right\rfloor=16$$ and hence obtained the first set of solutions.
To find the region where all of the coordinates of that region satisfy both conditions $$\left\lfloor x+\frac{7}{3} \right\rfloor=4$$ and $$\left\lfloor x-\frac{9}{4} \right\rfloor=0$$ simultaneously, we do as follows:
[TABLE="class: grid, width: 500"]
[TR]
[TD]$$\left\lfloor x+\frac{7}{3} \right\rfloor=4$$[/TD]
[TD]$$\left\lfloor x-\frac{9}{4} \right\rfloor=0$$[/TD]
[/TR]
[TR]
[TD]$$4\le x+\frac{7}{3}<5$$[/TD]
[TD] $$0\le x-\frac{9}{4}<1$$[/TD]
[/TR]
[TR]
[TD]$$\frac{5}{3}\le x< \frac{8}{3}$$[/TD]
[TD]$$\frac{9}{4}\le x< \frac{13}{4}$$[/TD]
[/TR]
[/TABLE]
Hence, the first solution set that we found is $$\frac{9}{4}\le x<\frac{8}{3}$$.
I repeat the process to find the second solution set by considering the negative value of $$x$$ and this leads me to assume the cases $$\left\lfloor x+\frac{7}{3} \right\rfloor=-5$$ and $$\left\lfloor x-\frac{9}{4} \right\rfloor=-9$$ then I would get the $$\left\lfloor x+\frac{7}{3}\right\rfloor^2+\left\lfloor x-\frac{9}{4}\right\rfloor=25-9=16$$ and hence get one of the set of solutions.
[TABLE="class: grid, width: 500"]
[TR]
[TD]$$\left\lfloor x+\frac{7}{3} \right\rfloor=-5$$[/TD]
[TD]$$\left\lfloor x-\frac{9}{4} \right\rfloor=-9$$[/TD]
[/TR]
[TR]
[TD]$$-5\le x+\frac{7}{3}<-4$$[/TD]
[TD] $$-9\le x-\frac{9}{4}<-8$$[/TD]
[/TR]
[TR]
[TD]$$-\frac{22}{3}\le x< -\frac{19}{3}$$[/TD]
[TD]$$-\frac{27}{4}\le x< -\frac{23}{4}$$[/TD]
[/TR]
[/TABLE]
This gives me the second solution set $$-\frac{27}{4}\le x<-\frac{19}{3}$$.
Therefore, $$-\frac{27}{4}\le x<-\frac{19}{3}$$ and $$\frac{9}{4}\le x<\frac{8}{3}$$ are the solution sets to this problem.