**Let's solve the given problem by factoring the expression on the right side and completing the missing part on the left side accordingly,**

We will then examine each of the algebraic expressions in both sides of the given equation __separately__,

__On the right side of the equation, the expression:__

$10a-90$__Let's now examine the expression on the left side of the equation:__

$?(18-2a)$Let's note that in this expression there is a factor (unknown) multiplying an expression in parentheses, therefore to understand what this factor is - we'll return to the expression on the right side and factor it using common factor extraction,

__In a routine manner__ - we'll look for __first __the largest common factor, __we'll do this separately__, for the numbers and for the letters:

__Let's start with the numbers:__

In the expression on the right side, there are the numerical coefficients - 90 and 10, let's note that the number 90 **is a multiple of** the number 10:

$90=9\cdot10$Therefore, the number 10 is **the largest common factor for the numbers**,

__Let's continue and examine the letters:__

Let's note that only the left term in the expression on the right side, meaning the term -

$10a$depends on $a$unlike the second term in this expression:

$-90$**which does not depend** on $a$

**Therefore, there is no common factor for these two terms (so we'll consider 1 as the common factor for the letters),**

__Let's summarize:__

The largest common factor **for numbers and letters together** is:

$10\cdot1\\
\downarrow\\
10$

**Let's continue then and factor** (using common factor extraction) the expression on the right side:

$\textcolor{red}{ 10a} \textcolor{blue}{-90 } \\
\underline{10}\cdot\textcolor{red}{a}+\underline{10}\cdot(\textcolor{blue}{ -9} )\\
\downarrow\\
\underline{10}\cdot(\textcolor{red}{a}\textcolor{blue}{ -9} )$

In the above expression, the operation is explained using colors and markings:

The common factor is highlighted __with an underline__, and the multipliers inside the parentheses are associated with the terms in the original expression using colors, __let's note that we referred in the factoring details above both to the sign of the common factor (in black) that we extracted as a multiplier outside the parentheses and to the signs of the terms in the original expression (in colors)__, it's not necessary to present this in stages as described above, one can (and should) jump directly to the factored form in the last line, but we must definitely refer to the aforementioned signs, since in each term the sign is an inseparable part of it,

**We can easily verify that this factoring is correct **__easily__ by opening the parentheses __using the distributive property __and verifying that __indeed we get back the original expression we factored __term by term, __it's advisable to do this while emphasizing the signs of the terms in the original expression and the sign (always optional) of the common factor.__

__Let's return____ to the given problem:__

We factored the expression on the right side of the given equation, let's apply this factoring in the equation itself:

$?(18-2a)=10a-90 \\
\downarrow\\
?(18-2a)=10(a-9)$

__Let's note that the algebraic expressions in parentheses on both sides of the equality____ are not identical,__

Therefore, **we'll examine the difference** between the expressions inside the parentheses, we want to bring the expression in parentheses on the right side to be __identical__ to the expression in parentheses on the left side, let's note that between the two terms in parentheses on the left side and between the two terms in the expression in parentheses on the right side **there is a clear proportion**:

$\frac{18}{-9}=\frac{-2a}{a}=-2$** And additionally** let's note that according to the multiplication rules:

$10=(-5)\cdot(-2)$Therefore, we can first rearrange (using the commutative property of addition) the expression in parentheses on the left side, in parallel we'll present the number 10 as a multiplication of numbers as mentioned above:

$10(a-9)\\
\downarrow\\
(-5)\cdot(-2)(-9+a)$__Let's emphasize again __that the sign preceding each term is an inseparable part of it, and **therefore when using the commutative property (in addition) we switched the places of the terms along with their preceding signs**.

**Let's continue** then and note that if we "insert" the factor $-2$into the parentheses, __we'll get that the expressions in parentheses on both sides of the equality will be identical __, we'll do this of course by applying the multiplication operation on the expression in parentheses,** using the distributive property:**

$x(y+z)=xy+xz$__Let's apply this in the expression we got in the last stage__, which is the expression on the right side in the given equation:

$(-5)\cdot(-2)(-9+a)\\
\downarrow\\
(-5)\big((-2)\cdot(-9)+(-2)\cdot a\big)\\
(-5)(18-2a)$In the last stage, we simplified the expression in parentheses,

**Now let's return again to the original problem and summarize the development stages**, we got that:

$?(18-2a)=10a-90 \\
\downarrow
?(18-2a)=10(a-9) \\
?(18-2a)=(-5)(-2)(a-9) \\
?(18-2a)=(-5)(18-2a) \\$

We got that the expressions in parentheses in the expressions on both sides of the equation are identical, therefore now we can__ easily__ complete the missing part** (the factor marked on the left side with a question mark)** and determine unequivocally that __the correct answer is answer C.__

__Important note:__

__This problem is not an equation-solving problem, but a problem of completing the missing part__, meaning - the solver is not required to find the value of the unknown (or unknowns) which when substituted in the equation will yield a true statement, **but to find the missing algebraic expressions **in the marked places and this **in order to get equality between the algebraic expressions on both sides of the equation (i.e., regardless of the value of the unknown)**, therefore in the above problem (for example) although clearly for:

$a=9$indeed there is equality between the sides of the equation,

__This information is of no use in order to answer what we were asked in the problem.__