not really, to factor those kinds of equations the middle 2 numbers need to add up into a certain number and and multiply into a certain number. In this case it would be 2x and 3x in the middle. After that you can just factor normally. box method can be used but it’s not the defacto method to solving simple quadratics! :D
I call it the box method I'm not sure what the official name is but basically it goes like this
First the basic step, finding two numbers that multiple to 6 and add to 5 which is 2 and 3
Then I make a box kind sort of like four square. The top left of the box is has X^2 and the top right has 2x, the bottom left has 3x and the bottom right has 6.
Then I try to find the factors by finding what they have in common from the collums and rows. For example the columns x^2 and 3x have X in common (technically it's 1x but in my country we write 1x and just x) then the next column 2x and 6 it's 2.
Moving onto rows x^2 and 2x have they have X in common and the next row 3x and 6 they have 2 in common.
Then I just gather my numbers making (x + 2) and (x+3), wish I could send images so you have a visial but eh
anyway my comment explained what i ment so oddly cause I was half asleep but my main point was in order to even solve/factor the equation always has to equal zero, if it doesn't you have to move some terms to make it equal 0 not 6
HE EXPLAINED THE EQUATION WRONG, TO FACTOR YOU GOTTA TUEN IT INTO FACTORED FORM FIRST ITS CURRENTLY IN VERTEX FORM SO YOU GOTTA USE THE BOX METHOD THEN SOLVE FROM THERE