The solutions to a certain quadratic equation are x = -4 and x = 3. Write the equation in standard form below.

[tex]y=a(x-x_1)(x-x_2)\\\\x_1=-4;\ x_2=3\ therefore\\\\y=a(x+4)(x-3)\\\\y=a(x^2+x-12)\\\\\boxed{y=ax^2+ax-12a}\ where\ a\in\mathbb{R}-\{0\}[/tex]