Question 1210612
The way I interpret the question, only the numbers given below the equation can be used to fill the blanks.
I am going to refer to the values to be filled as the constants A, B, C, D and E.
I will write the equation as
{{{sqrt(2x - 3) + Ax + B = Cx - D + E}}} .
It is not specified, but I am going to assume that each number can only be used once.
A linear equation is one where the variable appears multiplied by some coefficient(s), but never with an exponent, or in a square root.
The square root shown appears to make that equation not linear, so
I am also going to assume, that the equation to be made is to be "linear in disguise" because
{{{sqrt(2x-3)=0}}} for the value of {{{x}}} that satisfies {{{Ax + B = Cx - D + E}}} .
{{{sqrt(2x-3)=0}}} <--> {{{2x-3=0}}} <--> {{{x=3/2}}}
For {{{x=3/2}}} to be a solution of {{{Ax + B = Cx - D + E}}} , the values of the constants must be such that
{{{(3/2)A+B =(3/2)C-D+E}}}<-->{{{3A+2B=3C-2D+2E}}}<-->{{{3A-3C=2E-2D-2B}}}<-->{{{3(A-C)=2(E-D-B)}}} .
The values for A, B, C, D, and E must be all different, be in the list of numbers given, and the numbers given are such that {{{0<abs(A-C)<=20-8=12}}} .
The equation {{{3(A-C)=2(E-D-B)}}}<--> {{{A-C=2(E-D-B)/3}}}  tells us that
{{{A-C}}} must be a multiple of 2, and
{{{E-D-B}}} must be a multiple of 3.
From there on, we can guess and check to find suitable answers. There is a very large number of them.
For example, {{{system(E=20,D=8,B=9)}}} or {{{system(E=20,D=9,B=8)}}} give us
{{{E-D-B=20-8-9=3}}} and are part of some of the many possible answers.
Those choices make {{{A-C=2*3/3=2}}} ,
and leaves us the numbers 10, 11, 12, 13, 14, 15, 16, 17, 18, and 19 available to use as A and C.
Then, {{{system(A=12,C=10)}}} , {{{system(A=13,C=11)}}} ,  {{{system(A=14,C=12)}}} ,  {{{system(A=15,C=13)}}} , {{{system(A=16,C=14)}}} , {{{system(A=17,C=15)}}} , {{{system(A=18,C=16)}}}, {{{system(A=19,C=17)}}} could be used to complete an answer.
There are many other sets of values that are valid answers.
 
Let's verify if {{{system(A=12,B=8,C=10,D=9,E=20)}}} make {{{sqrt(2x-3)+12x+8= 10x-9+20}}} an equation "linear in disguise" that  has {{{x=3/2}}} as a solution.
Substituting {{{x=3/2}}} into {{{sqrt(2x-3)+12x+8= 10x-9+20}}} , I get
{{{sqrt(2(3/2)-3)+12(3/2)+8=10(3/2)-9+20}}}
{{{sqrt(3-3)+18+8=15-9+20}}} 
{{{sqrt(0)+26=6+20}}} 
{{{0+26=26}}}
{{{26=26}}}
 
For a more systematic approach, we could start with the possible values of {{{A-C}}} .
Knowing that {{{0<abs(A-C)<=20-8=12}}} , and that {{{A-C}}} is not zero,
the possible vales for {{{A-C}}} are -12, -10, -8, -6, -4, -2, 2, 4, 6, 8, 10, and 12.
That makes the possible values for {{{E-B-D=3(A-C)/2}}} -18,-15,-12, -9, -6 -3, 3, 6, 9, 12, 15, and 18. 
To get a longer list of possible solutions, we could start by using multiples of 3 for E, B, and D to make {{{E-B-D}}} a multiple of 3.
Thinking about modular arithmetic, for those familiar with it, we could make a very long list of options for E, B, and D that make {{{E-B-D}}} a multiple of 3
Then, we could weed out from those lists those with {{{E-B-D=0}}} and those with {{{abs(E-B-D)>18}}}, then we could find possible values of A and C, eliminating the ones where A or C was already used as E, B, or D.
We are still left with hundreds of different valid answers.