The solutions are 2 and -3 so we can write our equation as:
[tex]a\cdot(x-2)(x+3)=0\\\\a\cdot(x^2-2x+3x-6)=0\\\\a\cdot(x^2+x-6)=0\\\\ax^2+ax-a\cdot6=0[/tex]
and for a = 2 there will be:
[tex]ax^2+ax-a\cdot6=0\\\\
2x^2+2x-2\cdot6=0\\\\
2x^2+2x-12=0\quad|+12\\\\\boxed{2x^2+2x=12}[/tex]
Answer D.
