Algebra 2 (1st Edition)

We first check that the function is in the form $ax^2+bx+c$. Then, using the quadratic formula, we obtain: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ $$x=\frac{12\pm \sqrt{(-12)^2-4(3)(12)}}{2(3)}$$ $$x=2$$