[tex]a) \ f(x) = 2(x+5)(x-3)\\\\2(x+5)(x-3) = 0 \ \ /:2\\\\(x+5)(x-3) = 0\\\\x+5 = 0 \ \vee \ x -3 = 0\\\\x_1 = -5 \ \vee \ x_2 = 3\\\\x_1 + x_2 = -5+3\\\\\boxed{x_1+x_2 = -2}[/tex]
[tex]b) \ f(x) = 3(x-1)(x+4) \\\\3(x-1)(x+4) = 0 \ \ /:3\\\\(x-1)(x+4) = 0\\\\x-1 = 0 \ \vee \ x+4 = 0\\\\x_1 = 1 \ \vee \ x_2 = -4\\\\x_1+x_2 = 1-4\\\\\boxed{x_1+x_2 = -3}[/tex]
