Question 53222
Use the "ac" method.  Given an equation in the format {{{ax^2+bx+c=0}}} where {{{a<>1}}}, then {{{ac=3*18=54}}} in our case.  Now we need to find two numbers that when multiplied together is 54 and when added together is 'b' in the equation above, or 15 for us.  Make a table of the numbers:  2*27=54, but 2+27=29; 3*18=54, but 3+18=21; 6*9=54 and 6+9=15 --> we found them!  So, with the 6 and 9, we use them as coefficients and break up 'bx', thusly:  {{{3x^2+6x+9x+18=0}}}.  Now use grouping:  {{{(3x^2+6x)+(9x+18)=0}}}.  Now factor the two groupings:  {{{3x(x+2)+9(x+2)=0}}}.  The factor (x+2) is common now and we can factor that out to get {{{(x+2)(3x+9)=0}}}.  That means that either x+2=0 or 3x+9=0.  Solving both of them shows us that x=-2 or x=-3.