The sum of the maximum and minimum values of the function $$f(x) = |5x - 7| + [x^2 + 2x]$$ in the interval $$\left[\dfrac{5}{4}, 2\right]$$, where $$[t]$$ is the greatest integer $$\le t$$, is ______.

Given,

$$f(x)=|5x-7|+[x^2+2x],\qquad x\in\left[\frac54,2\right]$$

First analyze the modulus term.

$$5x-7=0\iff x=\frac75$$

Hence,

Case 1: $$\frac54\le x<\frac75$$

$$|5x-7|=7-5x$$

Case 2: $$\frac75\le x\le2$$

$$|5x-7|=5x-7$$

Now analyze

$$[x^2+2x]$$

Let

$$g(x)=x^2+2x$$

Since

$$g'(x)=2x+2>0,$$

$$g(x)$$ is increasing on the interval.

Now,

$$g\left(\frac54\right)=\frac{25}{16}+\frac52=\frac{65}{16}$$

$$=4.0625$$

and

$$g(2)=8$$

Hence,

$$[x^2+2x]$$

takes values

$$4,5,6,7,8$$

Now find transition points:

$$x^2+2x=5$$

$$x^2+2x-5=0$$

$$x=-1+\sqrt6$$

Similarly,

$$x^2+2x=6$$

$$x=-1+\sqrt7$$

and

$$x^2+2x=7$$

$$x=-1+2\sqrt2$$

Thus,

$$f(x)=7-5x+4=11-5x,\quad \frac54\le x<\frac75$$

This is decreasing.

Hence maximum in this part occurs at

$$x=\frac54$$

giving

$$f\left(\frac54\right)=11-\frac{25}{4}=\frac{19}{4}$$

Now for

$$\frac75\le x<-1+\sqrt6,$$

$$f(x)=5x-7+4=5x-3$$

which is increasing.

At

$$x=\frac75,$$

$$f\left(\frac75\right)=4$$

Next intervals:

$$f(x)=5x-2,\quad -1+\sqrt6\le x<-1+\sqrt7$$

$$f(x)=5x-1,\quad -1+\sqrt7\le x<-1+2\sqrt2$$

$$f(x)=5x,\quad -1+2\sqrt2\le x<2$$

and at

$$x=2,$$

$$f(2)=|10-7|+[8]=3+8=11$$

Hence minimum value is

$$4$$

and maximum value is

$$11$$

Therefore,

$$4+11=15$$

Hence, the required sum is

$$\boxed{15}$$.