According to the distributive law of multiplication,
a(b+c)=ab+ac
Therefore
(a+b)(c+d)=(a+b)c+(a+b)d=ac+bc+ad+bd
(x+3)(2x-1)
=(x+3)2x+(x+3)(-1)
=2x^2+6x-x-3
=2x^2+5x-3
(x+3)(2x+1)+3=0
(x+3)2x+(x+3)+3=0
2x^2+6x+x+3+3=0
2x^2+7x+6=0
(x+2)(2x+3)=0
x+2=0 or 2x+3=0
x=-2 or x=-3/2