Solve the Quadratic Equation: x²-3x+2=0 for X

Let's solve the given equation:

$x^2-3x+2=0$

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 will look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=2\\ m+n=-3\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we are looking for must yield a positive result, therefore we can conclude that both numbers have the same sign, according to multiplication rules, and now we'll remember that the possible factors of 2 are 2 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are identical will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-2\\ n=-1 \end{cases}$

therefore we will factor the expression on the left side of the equation to:

$x^2-3x+2=0 \\ \downarrow\\ (x-2)(x-1)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown on one side:

$x-2=0\\ \boxed{x=2}$

or:

$x-1=0\\ \boxed{x=1}$

Let's summarize then the solution of the equation:

$x^2-3x+2=0 \\ \downarrow\\ (x-2)(x-1)=0 \\ \downarrow\\ x-2=0\rightarrow\boxed{x=2}\\ x-1=0\rightarrow\boxed{x=1}\\ \downarrow\\ \boxed{x=2,1}$

Therefore the correct answer is answer B.