Using the following formula:
V2 = ((D-C1)/(C2-D))*V1
It is possible to estimate the amount of higly concentrated V2 with value C2 that has to be added to the base solution in order to increase the value C1 of that solution to D.
Example:
D = 50
C1 = 5
C2 = 500
V1 = 10
V2 = ((50-5)/(500-50))*10
Result: V2 = 1
So my total volume would be 11 (V1+V2) with concentration D = 50.
I need to increase 2 different substances A and B in one base solution, while using 2 highly concentrated substances in order to do that.
Let assume:
V1 = 10
DA = 50 - concentration of Substance A to reach
DB = 2000 - concentration of Substance B to reach
C2A = 500
C3B = 20000
V2 = ?
V3 = ?
I can't adapt the formula above to solve this problem: I need to add 2 different substances to base solution, but after estimating V2 and V3 and adding it to V1. It will dilute my correctly estimated concentrations.
Let's start by looking into the first problem in more detail first; even though you already have a formula for finding the desired volume $V_2$.
Since concentration is defined as quantity of the interesting substance divided by volume of the solution, our initial solution with volume $V_1$ contains $V_1\times C_1$ quantity of that substance. Likewise, volume $V_2$ of the highly-concentrated solution contains $V_2\times C_2$ quantity of that same substance. If we mix the two, our solution will have volume $(V_1+V_2)$ and contain $(V_1\times C_1+V_2\times C_2)$ of the substance and thus its concentration will be $$\frac{V_1\times C_1+V_2\times C_2}{V_1+V_2}$$
Setting this equal to the desired concentration $D$, and rearranging the terms gives us $$\begin{array}{rcl} \frac{V_1\times C_1+V_2\times C_2}{V_1+V_2} & = & D\\ V_1\times C_1+V_2\times C_2 & = & D\times(V_1+V_2)\\ V_2\times(C_2-D) & = & V_1\times(D-C_1)\\ V_2 & = & \frac{D-C_1}{C_2-D}\times V_1\\ \end{array}$$ Exactly the formula you have!
Now, the same idea can be used in the other case:
Once we mix them, the resulting mixture will have volume $(V_1+V_2+V_3)$ and contain $V_2\times C_{2A}$ of substance $A$ and $V_3\times C_{3B}$ of substance $B$. Thus, the concentration of $A$ will be $$\frac{V_2\times C_{2A}}{V_1+V_2+V_3}$$ and similarly for $B$. Since we are given the desired concentrations of $A$ and $B$, we thus need to have
$$\begin{array}{rlc} \frac{V_2\times C_{2A}}{V_1+V_2+V_3} & = & D_A \\ \frac{V_3\times C_{3B}}{V_1+V_2+V_3} & = & D_B \\ \end{array}$$
Again, a little bit of tidy up turns this into: $$\begin{array}{rlc} \underline{V_2}\times(C_{2A}-D_A) - \underline{V_3}\times D_A & = & D_A\times V_1 \\ \underline{V_3}\times(C_{3B}-D_B) - \underline{V_2}\times D_B & = & D_B\times V_1 \\ \end{array}$$ The underlined terms are the only unknown ones, so we have a system of two linear equations with two unknowns.
If you have a problem with specific values, it's usually easier to plug the values into the system now, rather than trying to come up (and remember!) with the general formula. For example, in the example provided, this system would look as follows:
$$\begin{array}{rlc} 450\times V_2 - 50\times V_3 & = & 500 \\ 18000\times V_3 - 2000\times V_2 & = & 20000 \\ \end{array}$$ Solving it yields $V_1=1.25$ and $V_2=1.25$.
Of course, one can obtain the fully general formulas for $V_2$ and $V_3$, they're just be somewhat unwieldy.
$$\begin{array}{rlc} V_2 & = & \frac{C_{3B}\times D_A}{(C_{3B}-D_B)\times(C_{2A}-D_A)-D_A\times D_B}\times V_1 \\ V_3 & = & \frac{C_{2A}\times D_B}{(C_{3B}-D_B)\times(C_{2A}-D_A)-D_A\times D_B}\times V_1 \\ \end{array}$$
Note that the thought process is quite general and can be applied to many other scenarios; such as when the initial mixture already contains some quantities of substances $A$ and $B$ - one still just needs to compute the total volume and the total quantities of substances $A$ and $B$ (and any others, if there are more of them) and build the system of equations. Of course, the more complicated the setting, the more complicated the resulting general formula would be.