Solve x²-7x+21=3x: Finding the Value of X

$x^2-7x+21=3x$

Solve the following problem:

Let's solve the following equation:

$x^2-7x+21=3x$

$$Let's begin by reorganizing the equation and combining like terms:

$x^2-7x+21=3x \\ x^2-7x+21-3x=0\\ x^2-10x+21=0$

Note that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

We'll look for a pair of numbers whose product is the constant term and whose sum is the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=21\\ m+n=-10\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for must yield a positive result. Therefore we can conclude that both numbers must have the same signs, according to multiplication rules. The possible factors of 21 are 7 and 3 or 21 and 1. Fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have equal signs, leads us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases}m=-7 \\ n=-3\end{cases}$

We'll proceed to factor the expression on the left side of the equation to:

$x^2-10x+21=0 \\ \downarrow\\ (x-7)(x-3)=0$

Remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Subsequently we obtain two simple equations and solve them by isolating the variable in each one:

$x-7=0\\ \boxed{x=7}$

or:

$x-3=0\\ \boxed{x=3}$

Let's summarize the solution of the equation:

$x^2-7x+21=3x \\ x^2-10x+21=0 \\ \downarrow\\ (x-7)(x-3)=0 \\ \downarrow\\ x-7=0\rightarrow\boxed{x=7}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=7,3}$

Therefore the correct answer is answer C.