To factorise and solve this equation we must put it into two brackets, instead of one, each with a single x term. As the is an x^2 term we know there must be an x in each bracket Because the number term at the end is negative we know one bracket must have a plus, and the other a negative. To figure out which numbers go in each bracket we must find two numbers that when multiplied together make -18 and when added/subtracted from each other make +3. The multiples of 18 are 1 and 18, 2 and 9 and 3 and 6. Only 3 and 6 can make the +3x we need. In order to make +3 we must have only bracket with -3 and the other with +6. This means we have factorised the equation to (x-3)(x+6)=0. To then solve the equations for the solutions to x we must put each bracket equal to 0 as that is the only way to produce 0 when two things are multiplied together. So, if x+6=0, then x=-6, and when x-3=0, x must be +3.