Processing Math: Done

From Förberedande kurs i matematik 1

Because both terms, x(x+3) and x(2x−9), contain the factor x, we can take out x from the left-hand side and collect together the remaining expression,

x(x+3)−x(2x−9)=x(x+3)−(2x−9)=x(x+3−2x+9)=x(−x+12).

The equation is thus

x(−x+12)=0

and we obtain directly that the equation is satisfied if either x or −x+12 is zero. The solutions to the equation are therefore x=0 and x=12.

Here, it can be worth checking that x=12 is a solution (the case x=0 is obvious)

LHS=12(12+3)−12(212−9)=215−1215=0=RHS.